Ý tưởng là xem phần còn lại của mọi phần tử khi chia cho 3. Một tập hợp các phần tử chỉ có thể tạo thành một nhóm nếu tổng các phần tử còn lại của chúng là bội số của 3.
Ví dụ. 8, 4, 12. Bây giờ, phần còn lại lần lượt là 2, 1 và 0. Điều này có nghĩa là 8 cách bội số 3 giây [6] 2 khoảng cách, 4 cách bội số 3 giây [3] 1 khoảng cách và 12 cách bội số 0. Vì vậy, chúng ta có thể viết tổng dưới dạng 8 [có thể viết là 6+2], 4 [có thể viết là 3+1] và 12 [có thể viết là 12+0]. Bây giờ tổng của 8, 4 và 12 có thể được viết là 6+2+3+1+12+0. Bây giờ, 6+3+12 sẽ luôn chia hết cho 3 vì tất cả các số hạng đều là bội của 3. Bây giờ, chúng ta chỉ cần kiểm tra xem 2+1+0 [số dư] có chia hết cho 3 hay không để tổng có chia hết cho 3.
Vì nhiệm vụ là liệt kê các nhóm nên chúng ta đếm tất cả các phần tử có số dư khác nhau.
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.Thực hiện
C++
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.791. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.801. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.821. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.84 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.85
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.01. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.11. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.31. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.51. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.71. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.101. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.121. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.141. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.161. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.19
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7921. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.793
_______09____3795
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7981. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.800
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8021. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.806
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.808_______09____3810
_______09____3812
_______09____3814
_______09____3816
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.818_______09____3820
_______09____3822
_______09____3824
_______09____3826
_______09____3828
_______09____3830
_______09____3832
_______09____3828
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.836_______09____3838
_______09____3840
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8431. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8451. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8471. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8511. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8541. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8551. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8551. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.011. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.021. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.031. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.061. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.081. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.09C
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.101. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.801. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.13
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.141. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.151. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.31. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.51. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.71. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.19
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7921. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.793
_______09____3795
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7981. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.800
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.421. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.806
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.48_______09____3814
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.52_______09____3820
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.56_______09____3826
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.60_______09____3832
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.64_______09____3838
_______09____3840
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8431. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.731. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8471. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8511. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8541. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8551. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8551. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.881. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.891. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.901. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.911. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.061. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844Java
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.961. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.801. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.811. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.99 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.141. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.15
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.31. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.51. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.71. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1211. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.21. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1301. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.133
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1361. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1381. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.793
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7951. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1441. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1471. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1481. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1491. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.150
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.421. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1541. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1601. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.162
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.481. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1681. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1701. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.172
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.521. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1761. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1821. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1861. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.138
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1761. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1821. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1861. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.138
_______3799____060
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1541. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1821. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1581. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1861. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.162
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.641. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1681. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1701. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1701. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.172
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.840________ 3799 ________ 3842 ________ 3843
_______09____3844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7939 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7940 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7941 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79421. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79481. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79511. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1491. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79601. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1301. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79641. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7966____0011. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79681. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7969_______09____3844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844Python3
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79731. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79741. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7975
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79761. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79771. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79781. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7979 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7980
_______09____37982
_______09____37984
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7986_______09____37988
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79901. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7992______37993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79941. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80021. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.125
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.9____38008_______09____38010
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80131. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8014 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80151. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.891. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80181. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80201. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8021 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1491. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000______380241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.159
_______09____38028
_______09____38030
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80351. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1601. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8046
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.9____38048_______09____38050
_______09____38052
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8057______11591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80571. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000
_______09____38065
_______09____38067
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000__1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178
_______09____38092
_______09____38094
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8039_______11591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000__1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.178
_______09____38119
_______09____38121
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80351. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80391. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8042 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80881. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8089 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8046
_______09____38119
_______09____38149
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80041. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8057______11251. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80571. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8038 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80571. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8000
_______09____38166
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.842 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8004
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81701. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79941. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1491. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.126______11871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79561. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79601. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81841. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81861. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8187
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8188 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.891. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81901. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8192____02____38194
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8195C#
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81961. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.801. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.811. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8200
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.99 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1001. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.9_______09____38206
_______09____38208
_______09____38210
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.31. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82151. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.71. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82221. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.799____382241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.799____382281. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82311. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.21. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82341. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.133
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7931. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.792
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.799____382441. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8249_______1147____38251
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.799____382531. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82551. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8257
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82591. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82611. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82631. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.814
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8161. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8181. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.799____382711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8192____38273
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8221. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8241. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82791. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81921. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8281
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8281. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8301. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.82871. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7968____38289
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8281. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8361. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8295
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.840________ 3799 ________ 3842 ________ 3843
_______09____3844
_______09____3845
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7939 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7940 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7941 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83091. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.122 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.79481. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83181. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.2 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83211. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7991. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83231. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.011. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8325____37969_______09____3844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.844
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8330PHP
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83311. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83321. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8333
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.01. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.11. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8336 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.3____383381. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1261. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83401. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.80461. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.81. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.101. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.121. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.141. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.161. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8352 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.7993____383541. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8355
_______09____3793
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8359 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8360
_______09____38362
_______09____38364
1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.91. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.797______0891. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8368 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83601. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8368 1. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83711. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83401. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.1381. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.83681. Hash all elements in a count array based on remainder, i.e,
for all elements a[i], do c[a[i]%3]++;
2. Now c[0] contains the number of elements which when divided
by 3 leave remainder 0 and similarly c[1] for remainder 1
and c[2] for 2.
3. Now for group of 2, we have 2 possibilities
a. 2 elements of remainder 0 group. Such possibilities are
c[0]*[c[0]-1]/2
b. 1 element of remainder 1 and 1 from remainder 2 group
Such groups are c[1]*c[2].
4. Now for group of 3,we have 4 possibilities
a. 3 elements from remainder group 0.
No. of such groups are c[0]C3
b. 3 elements from remainder group 1.
No. of such groups are c[1]C3
c. 3 elements from remainder group 2.
No. of such groups are c[2]C3
d. 1 element from each of 3 groups.
No. of such groups are c[0]*c[1]*c[2].
5. Add all the groups in steps 3 and 4 to obtain the result.8375