Given an array of integers, write a function that returns true if there is a triplet [a, b, c] that satisfies a2 + b2 = c2.
Example:
Input: arr[] = {3, 1, 4, 6, 5}
Output: True
There is a Pythagorean triplet [3, 4, 5].Input: arr[] = {10, 4, 6, 12, 5}
Output: False
There is no Pythagorean triplet.
Method 1 [Naive]
A simple solution is to run three loops, three loops pick three array elements, and check if the current three elements form a Pythagorean Triplet.
Below is the implementation of the above idea :
C++
#include
using namespace std;
bool isTriplet[int ar[], int n]
{
for [int
i = 0; i < n; i++] {
for [int j = i + 1; j < n; j++] {
for [int k = j + 1; k < n; k++] {
int x = ar[i] * ar[i], y = ar[j] * ar[j], z = ar[k] * ar[k];
if [x == y + z || y == x + z || z == x + y]
return true;
}
}
}
return false;
}
int main[]
{
int ar[] = { 3, 1, 4, 6, 5 };
int ar_size = sizeof[ar] / sizeof[ar[0]];
isTriplet[ar, ar_size] ? cout
Javascript
function isTriplet[ ar, n]
{
for [let i = 0; i < n; i++] {
for [let j = i + 1; j < n; j++] {
for [let k = j + 1; k < n; k++] {
let x = ar[i] * ar[i], y = ar[j] *
ar[j], z = ar[k] * ar[k];
if [x == y + z || y == x + z ||
z == x + y]
return true;
}
}
}
return false;
}
let ar = [ 3, 1, 4, 6, 5 ];
let ar_size = ar.length;
if [isTriplet[ar, ar_size] == true]
document.write["Yes"];
else
document.write["No"];
Output:
Yes
The Time Complexity of the above solution is O[n3].
Auxiliary Space: O[1]
Method 2 [Use Sorting]
We can solve this in O[n2] time by sorting the array first.
1] Do the square of every element in the input array. This step takes O[n] time.
2] Sort the squared array in increasing order. This step takes O[nLogn] time.
3] To find a triplet [a, b, c] such that a2 = b2 + c2, do
following.
- Fix ‘a’ as the last element of the sorted array.
- Now search for pair [b, c] in subarray between the first element and ‘a’. A pair [b, c] with a given sum can be found in O[n] time using the meet in middle algorithm discussed in method 1 of this post.
- If no pair is found for current ‘a’, then move ‘a’ one position back and repeat step 3.2.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
#include
#include
using namespace std;
bool isTriplet[int
arr[], int n]
{
for [int i = 0; i < n; i++]
arr[i] = arr[i] * arr[i];
sort[arr, arr + n];
for [int i = n - 1; i >= 2; i--] {
int l = 0;
int r = i - 1;
while [l < r] {
if [arr[l] + arr[r] == arr[i]]
return true;
[arr[l] + arr[r] < arr[i]] ? l++ : r--;
}
}
return false;
}
int
main[]
{
int arr[] = { 3, 1, 4, 6, 5 };
int arr_size = sizeof[arr] / sizeof[arr[0]];
isTriplet[arr, arr_size] ? cout = 2; i--] {
int l = 0;
int
r = i - 1;
while [l < r] {
if [arr[l] + arr[r] == arr[i]]
return true;
if [arr[l] + arr[r] < arr[i]]
l++;
else
r--;
}
}
return false;
}
public static void Main[]
{
int[] arr = { 3, 1, 4, 6, 5 };
int arr_size = arr.Length;
if
[isTriplet[arr, arr_size] == true]
Console.WriteLine["Yes"];
else
Console.WriteLine["No"];
}
}
PHP
Javascript
function isTriplet[arr , n]
{
for [i = 0; i < n; i++]
arr[i] = arr[i] * arr[i];
arr.sort[[a,b]=>a-b];
for [i = n - 1; i >= 2; i--]
{
var l = 0;
var r = i - 1;
while [l < r]
{
if [arr[l] + arr[r] == arr[i]]
return true;
if [arr[l] + arr[r] < arr[i]]
l++;
else
r--;
}
}
return
false;
}
var arr = [ 3, 1, 4, 6, 5 ];
var arr_size = arr.length;
if [isTriplet[arr, arr_size] == true]
document.write["Yes"];
else
document.write["No"];
Output:
Yes
The time complexity of this method is O[n2].
Auxiliary Space: O[1]
Method 3: [Using Hashing]
The
problem can also be solved using hashing. We can use a hash map to mark all the values of the given array. Using two loops, we can iterate for all the possible combinations of a and b, and then check if there exists the third value c. If there exists any such value, then there is a Pythagorean triplet.
Below is the implementation of the above approach:
C++
#include
using namespace std;
bool
checkTriplet[int arr[], int n]
{
int maximum = 0;
for [int i = 0; i < n; i++] {
maximum = max[maximum, arr[i]];
}
int hash[maximum + 1] = { 0 };
for [int i = 0; i < n; i++]
hash[arr[i]]++;
for [int i = 1; i < maximum + 1; i++] {
if [hash[i] == 0]
continue;
for [int j = 1; j < maximum + 1; j++] {
if
[[i == j && hash[i] == 1] || hash[j] == 0]
continue;
int val = sqrt[i * i + j * j];
if [[val * val] != [i * i + j * j]]
continue;
if [val > maximum]
continue;
if [hash[val]] {
return true;
}
}
}
return false;
}
int main[]
{
int arr[] = { 3, 2, 4, 6, 5 };
int
n = sizeof[arr] / sizeof[arr[0]];
if [checkTriplet[arr, n]]
cout maximum]:
continue
if [hash[val]]:
return True
return False
arr = [3, 2, 4, 6, 5]
n = len[arr]
if [checkTriplet[arr, n]]:
print["Yes"]
else:
print["No"]
C#
using System;
class GFG
{
static bool checkTriplet[int []arr, int n]
{
int maximum = 0;
for [int i = 0; i < n; i++]
{
maximum = Math.Max[maximum, arr[i]];
}
int []hash = new int[maximum + 1];
for [int i = 0; i < n; i++]
hash[arr[i]]++;
for [int i = 1; i < maximum + 1; i++]
{
if [hash[i] == 0]
continue;
for [int j = 1; j < maximum + 1; j++]
{
if [[i == j && hash[i] == 1] || hash[j] == 0]
continue;
int val = [int] Math.Sqrt[i * i + j * j];
if [[val * val] != [i * i + j * j]]
continue;
if [val > maximum]
continue;
if [hash[val] == 1]
{
return true;
}
}
}
return false;
}
public static void Main[String[] args]
{
int []arr = { 3, 2, 4, 6, 5 };
int n = arr.Length;
if [checkTriplet[arr, n]]
Console.Write["Yes"];
else
Console.Write["No"];
}
}
Javascript
function checkTriplet[arr , n] {
var maximum = 0;
for
[i = 0; i < n; i++] {
maximum = Math.max[maximum, arr[i]];
}
var hash = Array[maximum + 1].fill[0];
for [i = 0; i < n; i++]
hash[arr[i]]++;
for [i = 1; i < maximum + 1; i++] {
if [hash[i] == 0]
continue;
for [j = 1; j < maximum + 1; j++] {
if [[i == j && hash[i] == 1] || hash[j] == 0]
continue;
var val = parseInt[ Math.sqrt[i * i + j * j]];
if [[val * val] != [i * i + j * j]]
continue;
if [val > maximum]
continue;
if
[hash[val] == 1] {
return true;
}
}
}
return false;
}
var arr = [ 3, 2, 4, 6, 5 ];
var n = arr.length;
if [checkTriplet[arr, n]]
document.write["Yes"];
else
document.write["No"];
Thanks to
Striver for suggesting the above approach.
Time Complexity: O[ max * max ], where max is the maximum most element in the array.
Auxiliary Space: O[max]
Method -4:Using STL
Approach:
The problem can be solved using ordered maps and unordered maps. There is no need to store the elements in an ordered manner so implementation by an unordered map is faster. We can use the unordered map to mark all the values of the given array. Using two loops, we can iterate for all the possible combinations of a and b, and then check if there exists the third value c. If there exists any such value, then there is a Pythagorean triplet.
Below is the implementation of the above approach:
C++
#include
using namespace std;
bool checkTriplet[int
arr[], int n]
{
unordered_map umap;
for [int i = 0; i < n; i++]
umap[arr[i]] = umap[arr[i]] + 1;
for [int i = 0; i < n - 1; i++]
{
for [int j = i + 1; j < n; j++]
{
int p = sqrt[arr[i] * arr[i] + arr[j] * arr[j]];
float q
= sqrt[arr[i] * arr[i] + arr[j] * arr[j]];
if [p == q && umap[p] != 0]
return
true;
}
}
return false;
}
int main[]
{
int arr[] = { 3, 2, 4, 6, 5 };
int n = sizeof[arr] / sizeof[arr[0]];
if [checkTriplet[arr, n]]
cout