In max contiguous subarray in Python, you are given a list of integers[positive and negative], and you have to find a sub-list from the given list which has the maximum sum and print the sum.
For example: list = [1, 6, -7, 5]
now, possible contiguous lists are:
[1]
[1, 6]
[1, 6, -7]
[1, 6, -7, 5]
[6]
[6, -7]
[6, -7, 5]
[-7]
[-7, 5]
[5]
Now, the list with the highest sum is the 2nd list having 7 as the total sum.
Now, let’s code the max sub-array problem.
myList = [-1, 2, 4, -7, 5, -9]
listOfSubLists = []
for i in range[len[myList]+1]:
for j in range[i+1, len[myList]+1]:
listOfSubLists.append[myList[i:j]]
listOfAllSum = []
for i in listOfSubLists:
listOfAllSum.append[sum[i]]
max = -1
index = 0
for i in range[len[listOfAllSum]]:
if listOfAllSum[i]>max:
max = listOfAllSum[i]
index = i
print['List with maximum sum is: ', listOfSubLists[index]]
print['Maximum sum is: ', max] Max Contiguous Subarray
Given a list of integers, write a program to identify the contiguous sub-list that has the largest sum and print the sum. Any non-empty slice of the list with step size 1 can be considered as a contiguous sub-list.Input
The input will contain space-separated integers, denoting the elements of the list.Output
The output should be an integer.Explanation
For example, if the given list is [2, -4, 5, -1, 2, -3], then all the possible contiguous sub-lists will be,
[2]
[2, -4]
[2, -4, 5]
[2, -4, 5, -1]
[2, -4, 5, -1, 2]
[2, -4, 5, -1, 2, -3]
[-4]
[-4, 5]
[-4, 5, -1]
[-4, 5, -1, 2]
[-4, 5, -1, 2, -3]
[5]
[5, -1]
[5, -1, 2]
[5, -1, 2, -3]
[-1]
[-1, 2]
[-1, 2, -3]
[2]
[2, -3]
[-3]Among the above contiguous sub-lists, the contiguous sub-list [5, -1, 2] has the largest sum which is 6.
Sample Input 1
2 -4 5 -1 2 -3
Sample Output 1
6
Sample Input 2
-2 -3 4 -1 -2 1 5 -3
Sample Output 2
7
Max Contiguous Subarray
Given a list of integers, write a program to identify the contiguous sub-list that has the largest sum and print the sum. Any non-empty slice of the list with step size 1 can be considered as a contiguous sub-list.
Input
The input will contain space-separated integers, denoting the elements of the list.
Output
The output should be an integer.
Explanation
For example, if the given list is [2, -4, 5, -1, 2, -3], then all the possible contiguous sub-lists will be,
[2]
[2, -4]
[2, -4, 5]
[2, -4, 5, -1]
[2, -4, 5, -1, 2]
[-4]
[-4, 5]
[-4, 5, -1]
[-4, 5, -1, 2]
[5]
[5, -1]
[5, -1, 2]
[-1]
[-1, 2]
[2]
Among the above contiguous sub-lists, the contiguous sub-list [5, -1, 2] has the largest sum which is 6.
Sample Input 1
2 -4 5 -1 2 -3
Sample Output 1
6
Sample Input 2
-2 -3 4 -1 -2 1 5 -3
Sample Output 2
7
Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum.
Kadane’s Algorithm:
Initialize:
max_so_far = INT_MIN
max_ending_here = 0
Loop for each element of the array
[a] max_ending_here = max_ending_here + a[i]
[b] if[max_so_far < max_ending_here]
max_so_far = max_ending_here
[c] if[max_ending_here < 0]
max_ending_here = 0
return max_so_farExplanation:
The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array [max_ending_here is used for this]. And
keep track of maximum sum contiguous segment among all positive segments [max_so_far is used for this]. Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far
Lets take the example:
{-2, -3, 4, -1, -2, 1, 5, -3}
max_so_far = max_ending_here = 0
for i=0, a[0] = -2
max_ending_here = max_ending_here + [-2]
Set max_ending_here = 0 because max_ending_here < 0
for i=1, a[1] = -3
max_ending_here = max_ending_here + [-3]
Set max_ending_here = 0 because max_ending_here < 0
for i=2, a[2] = 4
max_ending_here = max_ending_here + [4]
max_ending_here = 4
max_so_far is updated to 4 because max_ending_here greater
than max_so_far which was 0 till now
for i=3, a[3] = -1
max_ending_here = max_ending_here + [-1]
max_ending_here = 3
for i=4, a[4] = -2
max_ending_here = max_ending_here + [-2]
max_ending_here = 1
for i=5, a[5] = 1
max_ending_here = max_ending_here + [1]
max_ending_here = 2
for i=6, a[6] = 5
max_ending_here = max_ending_here + [5]
max_ending_here = 7
max_so_far is updated to 7 because max_ending_here is
greater than max_so_far
for i=7, a[7] = -3
max_ending_here = max_ending_here + [-3]
max_ending_here = 4Program:
Python3
from math import inf
maxint=inf
def maxSubArraySum[a,size]:
max_so_far = -maxint - 1
max_ending_here =
0
for i in range[0, size]:
max_ending_here = max_ending_here + a[i]
if [max_so_far < max_ending_here]:
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
a =
[-13, -3, -25, -20, -3, -16, -23, -12, -5, -22, -15, -4, -7]
print ["Maximum contiguous sum is", maxSubArraySum[a,len[a]]]
Output:
Maximum contiguous sum is 7
Another approach:
Python3
def maxSubArraySum[a,size]:
max_so_far = a[0]
max_ending_here = 0
for i in range[0, size]:
max_ending_here = max_ending_here + a[i]
if max_ending_here < 0:
max_ending_here = 0
elif
[max_so_far < max_ending_here]:
max_so_far = max_ending_here
return max_so_far
Time Complexity: O[n]
Algorithmic Paradigm: Dynamic Programming
Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative.
Python3
def maxSubArraySum[a,size]:
max_so_far =a[0]
curr_max = a[0]
for i in range[1,size]:
curr_max = max[a[i], curr_max + a[i]]
max_so_far = max[max_so_far,curr_max]
return max_so_far
a =
[-2, -3, 4, -1, -2, 1, 5, -3]
print["Maximum contiguous sum is" , maxSubArraySum[a,len[a]]]
Output:
Maximum contiguous sum is 7
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.
Python3
from sys import maxsize
def maxSubArraySum[a,size]:
max_so_far = -maxsize - 1
max_ending_here = 0
start = 0
end = 0
s = 0
for i in range[0,size]:
max_ending_here += a[i]
if
max_so_far < max_ending_here:
max_so_far = max_ending_here
start = s
end = i
if max_ending_here < 0:
max_ending_here = 0
s = i+1
print ["Maximum contiguous sum is %d"%[max_so_far]]
print ["Starting Index %d"%[start]]
print ["Ending Index %d"%[end]]
a =
[-2, -3, 4, -1, -2, 1, 5, -3]
maxSubArraySum[a,len[a]]
Output:
Maximum contiguous sum is 7 Starting index 2 Ending index 6
Kadane’s Algorithm can be viewed both as a greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 [Greedy Part]. This is because continuing with a negative sum is way more worse than restarting with a new range. Now it can also be viewed as a DP, at each stage we have 2 choices: Either take the current element and continue with previous sum OR restart a new range. These both choices are being taken care of in the implementation.
Time Complexity: O[n]
Auxiliary Space: O[1]
Now try the below question
Given an array of integers [possibly some
elements negative], write a C program to find out the *maximum product* possible by multiplying ‘n’ consecutive integers in the array where n ≤ ARRAY_SIZE. Also, print the starting point of the maximum product subarray.
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