PYTHON

find maximum sum in array of contiguous subarrays

#include <iostream>

using namespace std;

int main(){
    //Input Array
    int n;
    cin >> n;
    int arr[n];
    for(int i =0;i< n;i++){
    cin >> arr[i];
    }

    int currentSum = 0;
    int maxSum = INT_MIN;
    //algo
    for (int i = 0; i < n; i++)
    {
        currentSum += arr[i];
        if (currentSum <0)
        {
            currentSum = 0;
        }
        maxSum = max(maxSum, currentSum);
    }
    cout << maxSum << endl;

    return 0;
}

Maximum Subarray sum

 def maxSubArray(self, nums: List[int]) -> int:
        curr_best = overall_best =  nums[0]
        for i in range(1,len(nums)):
            if curr_best>=0:
                curr_best = curr_best + nums[i]
            else:
                curr_best = nums[i]
            if curr_best > overall_best:
                overall_best = curr_best
        return overall_best

max subsequence sum in array

def max_subarray(numbers):
    """Find the largest sum of any contiguous subarray."""
    best_sum = 0  # or: float('-inf')
    current_sum = 0
    for x in numbers:
        current_sum = max(0, current_sum + x)
        best_sum = max(best_sum, current_sum)
    return best_sum

maximum sum subarray

public static int SumArray()
{
    var arr = new int[]{ -2, -4, -5, -6, -7, -89, -56 };
    var sum = 0;
    var max = arr[0];
    foreach (var item in arr)
    {
        sum += item;
      // sum = Math.Max(sum,0); resting here will not give  expected output
        max = Math.Max(sum,max);
        sum = Math.Max(sum,0);
    }
    return max;
}

maximum subarray

const maxSubArray = (arr) => {
    let currSum =  getSum(arr, 0, 4);
    let maxSum = 0
    for(let i = 1; i < arr.length - 4; i++){
       currSum =  currSum - arr[i - 1];
       currSum = currSum + arr[i + 3];
        maxSum = Math.max(currSum, maxSum);
    }
    return maxSum;
}

const getSum = (arr, start, end) => {
    let sum = 0;
    for(let i = start; i < end; i++){
        sum += arr[i]
    }
    
    console.log(sum);
    return sum;
}

How to find the maximum subarray sum in python?

"""
Implementation of "kadane's algorithm".
It returns the maximum subarray sum for
a given bigger array.
Time complexity: O(n)
Space complexity: O(1)
"""

def kadane_algorithm(array):
    running_sum = array[0]
    max_sum = array[0]
    for idx in range(1, len(array)):
        current_val = array[idx]
        running_sum = max(current_val, running_sum+current_val)
        max_sum = max(max_sum, running_sum)
    return max_sum

print(kadane_algorithm([1, 2, 3, 5, -1]))  # 11 = 1 + 2 + 3 + 5

Maximum subarray sum

var maxSequence = function(arr){
  var min = 0, ans = 0, i, sum = 0;
  for (i = 0; i < arr.length; ++i) {
    sum += arr[i];
    min = Math.min(sum, min);
    ans = Math.max(ans, sum - min);
  }
  return ans;
}

print maximum sum subarray

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n;
	cin>>n;
	int a[n];
	for(int  i = 0;i<n;i++){
		cin>>a[i];
	}
	int s = 0;
	int mx = INT_MIN;
	for(int  i =0;i<n;i++){
		s+=a[i];
		if(s<0){
			s=0;
		}
		mx = max(s,mx);
	}
	cout<<mx;
	return 0;
}

Largest sum contiguous subarray

// Java program to print largest 
// contiguous array sum
class GFG {
  
    static void maxSubArraySum(int a[], int size)
    {
        int max_so_far = Integer.MIN_VALUE,
        max_ending_here = 0,start = 0,
        end = 0, s = 0;
  
        for (int i = 0; i < size; i++) 
        {
            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;
            }
        }
        System.out.println("Maximum contiguous sum is " 
                           + max_so_far);
        System.out.println("Starting index " + start);
        System.out.println("Ending index " + end);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int n = a.length;
        maxSubArraySum(a, n);
    }
}
  
// This code is contributed by  prerna saini

maximum subarray

// A Divide and Conquer based Java
// A for maximum subarray sum
// problem
import java.util.*;
 
class GFG {
 
    // Find the maximum possible sum in arr[]
    // such that arr[m] is part of it
    static int maxCrossingSum(int arr[], int l, int m,
                              int h)
    {
        // Include elements on left of mid.
        int sum = 0;
        int left_sum = Integer.MIN_VALUE;
        for (int i = m; i >= l; i--) {
            sum = sum + arr[i];
            if (sum > left_sum)
                left_sum = sum;
        }
 
        // Include elements on right of mid
        sum = 0;
        int right_sum = Integer.MIN_VALUE;
        for (int i = m; i <= h; i++) {
            sum = sum + arr[i];
            if (sum > right_sum)
                right_sum = sum;
        }
 
        // Return sum of elements on left
        // and right of mid
        // returning only left_sum + right_sum will fail for
        // [-2, 1]
        return Math.max(left_sum + right_sum - arr[m],
                        Math.max(left_sum, right_sum));
    }
 
    // Returns sum of maximum sum subarray
    // in aa[l..h]
    static int maxSubArraySum(int arr[], int l, int h)
    {
          //Invalid Range: low is greater than high
          if (l > h)
              return Integer.MIN_VALUE;
        // Base Case: Only one element
        if (l == h)
            return arr[l];
 
        // Find middle point
        int m = (l + h) / 2;
 
        /* Return maximum of following three
        possible cases:
        a) Maximum subarray sum in left half
        b) Maximum subarray sum in right half
        c) Maximum subarray sum such that the
        subarray crosses the midpoint */
        return Math.max(
            Math.max(maxSubArraySum(arr, l, m-1),
                     maxSubArraySum(arr, m + 1, h)),
            maxCrossingSum(arr, l, m, h));
    }
 
    /* Driver program to test maxSubArraySum */
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 4, 5, 7 };
        int n = arr.length;
        int max_sum = maxSubArraySum(arr, 0, n - 1);
 
        System.out.println("Maximum contiguous sum is "
                           + max_sum);
    }
}
// This code is contributed by Prerna Saini

Maximum sum subarray of size ‘K’

l=[4,6,10,8,2,1]
ans=0
m=3
for i in range(len(l)+1):
   for j in range(i):
       if len(l[j:i])==m:
          ans=max(ans,sum(l[j:i]))
print(ans)