Maximum sum subarray having sum less than or equal to given sum

Given an array of non-negative integers and a sum. We have to find sum of the subarray having a maximum sum less than or equal to the given sum in the array. 
Note: Given array contains only non-negative integers.

Examples:

Input: arr[] = { 1, 2, 3, 4, 5 }
        sum = 11
Output: 10
Subarray having maximum sum is { 1, 2, 3, 4 }

Input: arr[] = { 2, 4, 6, 8, 10 }
        sum = 7
Output: 6
Subarray having maximum sum is { 2, 4 } or { 6 }

Naive Approach: We can find the maximum sum of the subarray by running two loops.

C++

// C++ program to find subarray having
// maximum sum less than or equal to sum
#include <bits/stdc++.h>
using namespace std;

// To find subarray with maximum sum
// less than or equal to sum
int findMaxSubarraySum(int arr[], int n, int sum)
{
    int result = 0;

    for (int i = 0; i < n; i++) {
        int currSum = 0;
        for (int j = i; j < n; j++) {
            currSum += arr[j];

            if (currSum < sum) {
                result = max(result, currSum);
            }
        }
    }
    return result;
}

// Driver program to test above function
int main()
{
    int arr[] = { 6, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 20;

    cout << findMaxSubarraySum(arr, n, sum);

    return 0;
}

Java

import java.io.*;
import java.util.*;

public class Gfg {
    // To find subarray with maximum sum
    // less than or equal to sum
    static int findMaxSubarraySum(int[] arr, int n, int sum)
    {
        int result = 0;

        for (int i = 0; i < n; i++) {
            int currSum = 0;
            for (int j = i; j < n; j++) {
                currSum += arr[j];

                if (currSum < sum) {
                    result = Math.max(result, currSum);
                }
            }
        }
        return result;
    }

    public static void main(String[] args)
    {
        int[] arr = { 6, 8, 9 };
        int n = arr.length;
        int sum = 20;

        System.out.println(findMaxSubarraySum(arr, n, sum));
    }
}

Python3

# Python3 program to find subarray having
# maximum sum less than or equal to sum

# To find subarray with maximum sum
# less than or equal to sum
def findMaxSubarraySum(arr, n, sum):
    result = 0

    for i in range(n):
        currSum = 0
        for j in range(i, n):
            currSum += arr[j]

            if currSum < sum:
                result = max(result, currSum)
    return result

# Driver program to test above function
if __name__ == '__main__':
    arr = [6, 8, 9]
    n = len(arr)
    sum = 20

    print(findMaxSubarraySum(arr, n, sum))
# This code is contributed by Prajwal Kandekar

C#

using System;

class Gfg {
    // To find subarray with maximum sum
    // less than or equal to sum
    static int findMaxSubarraySum(int[] arr, int n, int sum)
    {
        int result = 0;

        for (int i = 0; i < n; i++) {
            int currSum = 0;
            for (int j = i; j < n; j++) {
                currSum += arr[j];

                if (currSum < sum) {
                    result = Math.Max(result, currSum);
                }
            }
        }
        return result;
    }

    public static void Main()
    {
        int[] arr = { 6, 8, 9 };
        int n = arr.Length;
        int sum = 20;

        Console.WriteLine(findMaxSubarraySum(arr, n, sum));
    }
}

Javascript

// To find subarray with maximum sum
// less than or equal to sum
function findMaxSubarraySum(arr, n, sum) {
    let result = 0;

    for (let i = 0; i < n; i++) {
        let currSum = 0;
        for (let j = i; j < n; j++) {
            currSum += arr[j];
            if (currSum < sum) {
                result = Math.max(result, currSum);
            }
        }
    }
    return result;
}

const arr = [6, 8, 9];
const n = arr.length;
const sum = 20;

console.log(findMaxSubarraySum(arr, n, sum));

Output

17

Time Complexity: O(N*N)
Auxiliary Space: O(1)

Efficient Approach: The subarray having maximum sum can be found by using a sliding window. If curr_sum is less than sum include array elements to it. If it becomes greater than sum removes elements from start in curr_sum.  (This will work only in the case of non-negative elements.)

Implementation:

C++

// C++ program to find subarray having
// maximum sum less than or equal to sum
#include <bits/stdc++.h>
using namespace std;

// To find subarray with maximum sum
// less than or equal to sum
int findMaxSubarraySum(int arr[], int n, int sum)
{
    // To store current sum and
    // max sum of subarrays
    int curr_sum = arr[0], max_sum = 0, start = 0;

    // To find max_sum less than sum
    for (int i = 1; i < n; i++) {

        // Update max_sum if it becomes
        // greater than curr_sum
        if (curr_sum <= sum)
            max_sum = max(max_sum, curr_sum);

        // If curr_sum becomes greater than
        // sum subtract starting elements of array
        while (start < i && curr_sum + arr[i] > sum) {
            curr_sum -= arr[start];
            start++;
        }

        // If cur_sum becomes negative then start new subarray
        if (curr_sum < 0)
        {
            curr_sum = 0;
        }

        // Add elements to curr_sum
        curr_sum += arr[i];

    }

    // Adding an extra check for last subarray
    if (curr_sum <= sum)
        max_sum = max(max_sum, curr_sum);

    return max_sum;
}

// Driver program to test above function
int main()
{
    int arr[] = {6, 8, 9};
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 20;

    cout << findMaxSubarraySum(arr, n, sum);

    return 0;
}

Java

// Java program to find subarray having
// maximum sum less than or equal to sum
public class Main {

    // To find subarray with maximum sum
    // less than or equal to sum
    static int findMaxSubarraySum(int arr[],
                             int n, int sum)
    {
    // To store current sum and
    // max sum of subarrays
    int curr_sum = arr[0], max_sum = 0, start = 0;

    // To find max_sum less than sum
    for (int i = 1; i < n; i++) {

        // Update max_sum if it becomes
        // greater than curr_sum
        if (curr_sum <= sum)
           max_sum = Math.max(max_sum, curr_sum);

        // If curr_sum becomes greater than
        // sum subtract starting elements of array
        while (curr_sum + arr[i] > sum && start < i) {
            curr_sum -= arr[start];
            start++;
        }

        // Add elements to curr_sum
        curr_sum += arr[i];
    }

    // Adding an extra check for last subarray
    if (curr_sum <= sum)
        max_sum = Math.max(max_sum, curr_sum);

    return max_sum;
    }

    // Driver program to test above function
    public static void main(String[] args)
    {
        int arr[] = { 6, 8, 9};
        int n = arr.length;
        int sum = 20;

        System.out.println(findMaxSubarraySum(arr, n, sum));
    }
}

Python3

# Python3 program to find subarray having
# maximum sum less than or equal to sum

# To find subarray with maximum sum
# less than or equal to sum
def findMaxSubarraySum(arr, n, sum):

    # To store current sum and
    # max sum of subarrays
    curr_sum = arr[0]
    max_sum = 0
    start = 0;

    # To find max_sum less than sum
    for i in range(1, n):

        # Update max_sum if it becomes
        # greater than curr_sum
        if (curr_sum <= sum):
            max_sum = max(max_sum, curr_sum)

        # If curr_sum becomes greater than sum
        # subtract starting elements of array
        while (curr_sum + arr[i] > sum and start < i):
            curr_sum -= arr[start]
            start += 1

        # Add elements to curr_sum
        curr_sum += arr[i]

    # Adding an extra check for last subarray
    if (curr_sum <= sum):
        max_sum = max(max_sum, curr_sum)

    return max_sum

# Driver Code
if __name__ == '__main__':
    arr = [6, 8, 9]
    n = len(arr)
    sum = 20

    print(findMaxSubarraySum(arr, n, sum))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find subarray
// having maximum sum less
//than or equal to sum
using System;

public class GFG
{

    // To find subarray with maximum
    // sum less than or equal
    // to sum
    static int findMaxSubarraySum(int []arr,
                             int n, int sum)
    {    // To store current sum and
    // max sum of subarrays
    int curr_sum = arr[0], max_sum = 0, start = 0;

    // To find max_sum less than sum
    for (int i = 1; i < n; i++) {

        // Update max_sum if it becomes
        // greater than curr_sum
        if (curr_sum <= sum)
           max_sum = Math.Max(max_sum, curr_sum);

        // If curr_sum becomes greater than
        // sum subtract starting elements of array
        while (curr_sum + arr[i] > sum && start < i) {
            curr_sum -= arr[start];
            start++;
        }

        // Add elements to curr_sum
        curr_sum += arr[i];
    }

    // Adding an extra check for last subarray
    if (curr_sum <= sum)
        max_sum = Math.Max(max_sum, curr_sum);

    return max_sum;
    }

    // Driver Code
    public static void Main()
    {
        int []arr = {6, 8, 9};
        int n = arr.Length;
        int sum = 20;

        Console.Write(findMaxSubarraySum(arr, n, sum));
    }
}

// This code is contributed by Nitin Mittal.

Javascript

<script>

// Javascript program to find subarray having
// maximum sum less than or equal to sum

// To find subarray with maximum sum
// less than or equal to sum
function findMaxSubarraySum(arr, n, sum)
{

    // To store current sum and
    // max sum of subarrays
    let curr_sum = arr[0], max_sum = 0,
        start = 0;

    // To find max_sum less than sum
    for(let i = 1; i < n; i++)
    {

        // Update max_sum if it becomes
        // greater than curr_sum
        if (curr_sum <= sum)
            max_sum = Math.max(max_sum, curr_sum);

        // If curr_sum becomes greater than
        // sum subtract starting elements of array
        while (curr_sum + arr[i] > sum && start < i)
        {
            curr_sum -= arr[start];
            start++;
        }

        // Add elements to curr_sum
        curr_sum += arr[i];
    }

    // Adding an extra check for last subarray
    if (curr_sum <= sum)
        max_sum = Math.max(max_sum, curr_sum);

    return max_sum;
}

// Driver code
let arr = [ 6, 8, 9 ];
let n = arr.length;
let sum = 20;

document.write(findMaxSubarraySum(arr, n, sum));

// This code is contributed by Mayank Tyagi

</script>

PHP

<?php
// PHP program to find subarray having
// maximum sum less than or equal to sum

// To find subarray with maximum sum
// less than or equal to sum
function findMaxSubarraySum(&$arr, $n, $sum)
{
    // To store current sum and
    // max sum of subarrays
    $curr_sum = $arr[0];
    $max_sum = 0;
    $start = 0;

    // To find max_sum less than sum
    for ($i = 1; $i < $n; $i++)
    {

        // Update max_sum if it becomes
        // greater than curr_sum
        if ($curr_sum <= $sum)
        $max_sum = max($max_sum, $curr_sum);

        // If curr_sum becomes greater than
        // sum subtract starting elements of array
        while ($curr_sum + $arr[$i] > $sum &&
                             $start < $i)
        {
            $curr_sum -= $arr[$start];
            $start++;
        }

        // Add elements to curr_sum
        $curr_sum += $arr[$i];
    }

    // Adding an extra check for last subarray
    if ($curr_sum <= $sum)
        $max_sum = max($max_sum, $curr_sum);

    return $max_sum;
}

// Driver Code
$arr = array(6, 8, 9);
$n = sizeof($arr);
$sum = 20;

echo findMaxSubarraySum($arr, $n, $sum);

// This code is contributed by ita_c
?>

Output

17

Time Complexity: O(n)
Auxiliary Space: O(1)

If an array with all types(positive, negative or zero) of elements is given, we can use prefix sum and sets and worst case time complexity for that will be O(n.log(n)). You can refer Maximum Subarray sum less than or equal to k using set article for more clarity of this method.