Given an array arr[] of N integers, your task is to find the maximum sum of values in a contiguous, nonempty subarray.

Examples:

Input: N = 8, arr[] = {-1, 3, -2, 5, 3, -5, 2, 2}
Output: 9
Explanation: The subarray with maximum sum is {3, -2, 5, 3} with sum = 3 - 2 + 5 + 3 = 9.

Input: N = 6, arr[] = {-10, -20, -30, -40, -50, -60}
Output: -10
Explanation: The subarray with maximum sum is {-10} with sum = -10

Approach: To solve the problem, follow the below idea:

To solve the problem, we can maintain a running sum and check whenever the running sum becomes negative, we can reset it to zero. This is because if we have a subarray with negative sum and then include more elements to it, it will only decrease the total sum. Instead, we can remove the subarray with negative sum to get a greater subarray sum. The maximum running sum will be our final answer.

Step-by-step algorithm:

Below is the implementation of the algorithm:

• Maintain a variable sum to keep track of the running sum.
• Maintain a variable max_sum to keep track of the maximum running sum encountered so far.
• Iterate over the input array and add the current element to sum.
• If the sum becomes greater than max_sum, update max_sum = sum.
• If sum becomes negative, update sum = 0.
• After iterating over the entire array, return max_sum as the final answer.

Below is the implementation of the algorithm:

#include <bits/stdc++.h>
#define ll long long
using namespace std;

// function to find the maximum subarray sum
ll maxSubarraySum(ll* arr, ll N)
{
    // variables to store the running sum and the maximum
    // sum
    ll sum = 0, max_sum = -1e9;
    for (int i = 0; i < N; i++) {
        sum += arr[i];
        max_sum = max(max_sum, sum);
        // If we encounter a subarray with negative sum,
        // remove the subarray from the current sum
        if (sum < 0)
            sum = 0;
    }
    return max_sum;
}

int main()
{
    // Sample Input
    ll N = 8;
    ll arr[N] = { -1, 3, -2, 5, 3, -5, 2, 2 };

    cout << maxSubarraySum(arr, N) << endl;
}

#include <bits/stdc++.h>

#define ll long long

using namespace std;

​

// function to find the maximum subarray sum

ll maxSubarraySum(ll* arr, ll N)

{

    // variables to store the running sum and the maximum

    // sum

    ll sum = 0, max_sum = -1e9;

    for (int i = 0; i < N; i++) {

        sum += arr[i];

        max_sum = max(max_sum, sum);

        // If we encounter a subarray with negative sum,

        // remove the subarray from the current sum

        if (sum < 0)

            sum = 0;

    }

    return max_sum;

}

​

int main()

{

    // Sample Input

    ll N = 8;

    ll arr[N] = { -1, 3, -2, 5, 3, -5, 2, 2 };

​

    cout << maxSubarraySum(arr, N) << endl;

}

import java.util.*;

public class MaxSubarraySum {
    // Function to find the maximum subarray sum
    static long maxSubarraySum(long[] arr, int N) {
        // Variables to store the running sum and the maximum sum
        long sum = 0, max_sum = Long.MIN_VALUE;
        for (int i = 0; i < N; i++) {
            sum += arr[i];
            max_sum = Math.max(max_sum, sum);
            // If we encounter a subarray with negative sum,
            // remove the subarray from the current sum
            if (sum < 0)
                sum = 0;
        }
        return max_sum;
    }

    public static void main(String[] args) {
        // Sample Input
        int N = 8;
        long[] arr = { -1, 3, -2, 5, 3, -5, 2, 2 };

        System.out.println(maxSubarraySum(arr, N));
    }
}

// This code is contributed by akshitaguprzj3

using System;

public class GFG{
    // Function to find the maximum subarray sum
    static long MaxSubarraySum(long[] arr, int N) {
        // Variables to store the running sum and the maximum sum
        long sum = 0, max_sum = long.MinValue;
        for (int i = 0; i < N; i++) {
            sum += arr[i];
            max_sum = Math.Max(max_sum, sum);
            // If we encounter a subarray with negative sum,
            // remove the subarray from the current sum
            if (sum < 0)
                sum = 0;
        }
        return max_sum;
    }

    public static void Main() {
        // Sample Input
        int N = 8;
        long[] arr = { -1, 3, -2, 5, 3, -5, 2, 2 };

        Console.WriteLine(MaxSubarraySum(arr, N));
    }
}

// This code is contributed by rohit singh

// Function to find the maximum subarray sum
function maxSubarraySum(arr) {
    // Variables to store the running sum and the maximum sum
    let sum = 0;
    let max_sum = -1e9;

    for (let i = 0; i < arr.length; i++) {
        sum += arr[i];
        max_sum = Math.max(max_sum, sum);

        // If we encounter a subarray with a negative sum,
        // reset the sum to 0
        if (sum < 0) {
            sum = 0;
        }
    }

    return max_sum;
}

// Sample Input
const N = 8;
const arr = [-1, 3, -2, 5, 3, -5, 2, 2];
console.log(maxSubarraySum(arr));

# Function to find the maximum subarray sum
def max_subarray_sum(arr):
    # Variables to store the running sum and the maximum sum
    sum_val = 0
    max_sum = float('-inf')
    for num in arr:
        sum_val += num
        max_sum = max(max_sum, sum_val)
        # If we encounter a subarray with negative sum,
        # reset the current sum to 0
        if sum_val < 0:
            sum_val = 0
    return max_sum

# Driver code
if __name__ == "__main__":
    # Sample Input
    arr = [-1, 3, -2, 5, 3, -5, 2, 2]
    print(max_subarray_sum(arr))

9

Time Complexity: O(N), where N is the size of the input array arr[].
Auxiliary Space: O(1)