Maximum sum subarray having sum less than or equal to given sum
Given an array of integers and a sum. We have to find sum of subarray having maximum sum less than or equal to given sum in array.
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 maximum sum of subarray by running two loops. But the time complecxity will be O(N*N).
Efficient Approach: The subarray having maximum sum can be found by using sliding window. If curr_sum is less than sum include array elements to it. If it becomes greater than sum remove elements from start in curr_sum.
C++
// CPP 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 (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 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[] = { 1, 2, 3, 4, 5 }; int n = arr.length; int sum = 11; System.out.println(findMaxSubarraySum(arr, n, sum)); } } |
Python3
# Python 3 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 = {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 // 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
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