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

09 Nov, 2021
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. But the time complexity will be O(N*N).
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.)
C++
Java
Python3
C#
PHP
Javascript
// 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 sumint 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 functionint main(){    int arr[] = {6, 8, 9};    int n = sizeof(arr) / sizeof(arr[0]);    int sum = 20;     cout << findMaxSubarraySum(arr, n, sum);     return 0;}
// Java program to find subarray having// maximum sum less than or equal to sumpublic 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[] = { 1, 2, 3, 4, 5 };        int n = arr.length;        int sum = 11;         System.out.println(findMaxSubarraySum(arr, n, sum));    }}
# 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 sumdef 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 Codeif __name__ == '__main__':    arr = [6, 8, 9]    n = len(arr)    sum = 20     print(findMaxSubarraySum(arr, n, sum)) # This code is contributed by# Surendra_Gangwar
// C# program to find subarray// having maximum sum less//than or equal to sumusing 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 = {1, 2, 3, 4, 5};        int n = arr.Length;        int sum = 11;         Console.Write(findMaxSubarraySum(arr, n, sum));    }} // This code is contributed by Nitin Mittal.
<?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 sumfunction 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?>
<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 sumfunction 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 codelet 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>
17
 
Time Complexity: O(N*log(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. 
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ukasp
SURENDRA_GANGWAR
mayanktyagi1709
shreyadubey1802
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subarray-sum
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