Maximum sum subarray having sum less than given sum using Set

Last Updated : 12 Jul, 2025

Given an array arr[] of length N and an integer K, the task is the find the maximum sum subarray with a sum less than K.
Note: If K is less than the minimum element, then return INT_MIN.

Examples:

Input: arr[] = {-1, 2, 2}, K = 4
Output: 3
Explanation:
The subarray with maximum sum which is less than 4 is {-1, 2, 2}.
The subarray {2, 2} has maximum sum = 4, but it is not less than 4.

Input: arr[] = {5, -2, 6, 3, -5}, K =15
Output: 12
Explanation:
The subarray with maximum sum which is less than 15 is {5, -2, 6, 3}.

Efficient Approach: Sum of subarray [i, j] is given by cumulative sum till j - cumulative sum till i of the array. Now the problem reduces to finding two indexes i and j, such that i < j and cum[j] - cum[i] are as close to K but lesser than it.
To solve this, iterate the array from left to right. Put the cumulative sum of i values that you have encountered till now into a set. When you are processing cum[j] what you need to retrieve from the set is the smallest number in the set which is bigger than or equal to cum[j] - K. This can be done in O(logN) using upper_bound on the set.

Below is the implementation of the above approach:

C++

// C++ program to find maximum sum
// subarray less than K

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

// Function to maximum required sum < K
int maxSubarraySum(int arr[], int N, int K)
{

    // Hash to lookup for value (cum_sum - K)
    set<int> cum_set;
    cum_set.insert(0);

    int max_sum = INT_MIN, cSum = 0;

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

        // getting cumulative sum from [0 to i]
        cSum += arr[i];

        // lookup for upperbound
        // of (cSum-K) in hash
        set<int>::iterator sit
            = cum_set.upper_bound(cSum - K);

        // check if upper_bound
        // of (cSum-K) exists
        // then update max sum
        if (sit != cum_set.end())

            max_sum = max(max_sum, cSum - *sit);

        // insert cumulative value in hash
        cum_set.insert(cSum);
    }

    // return maximum sum
    // lesser than K
    return max_sum;
}

// Driver code
int main()
{

    // initialise the array
    int arr[] = { 5, -2, 6, 3, -5 };

    // initialise the value of K
    int K = 15;

    // size of array
    int N = sizeof(arr) / sizeof(arr[0]);

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

    return 0;
}

Java

// Java program to find maximum sum
// subarray less than K
import java.util.*;
import java.io.*;

class GFG{
    
// Function to maximum required sum < K
static int maxSubarraySum(int arr[], int N, 
                          int K)
{
    
    // Hash to lookup for value (cum_sum - K)
    Set<Integer> cum_set = new HashSet<>();
    cum_set.add(0);
 
    int max_sum =Integer.MIN_VALUE, cSum = 0;
 
    for(int i = 0; i < N; i++)
    {
        
        // Getting cumulative sum from [0 to i]
        cSum += arr[i];
 
        // Lookup for upperbound
        // of (cSum-K) in hash
        ArrayList<Integer> al = new ArrayList<>();
        Iterator<Integer> it = cum_set.iterator(); 
        int end = 0;
        
        while (it.hasNext()) 
        { 
            end = it.next();
            al.add(end);
        }
        
        Collections.sort(al);
        int sit = lower_bound(al, cSum - K);
        
        // Check if upper_bound
        // of (cSum-K) exists
        // then update max sum
        if (sit != end)
            max_sum = Math.max(max_sum, 
                               cSum - sit);
 
        // Insert cumulative value in hash
        cum_set.add(cSum);
    }
 
    // Return maximum sum
    // lesser than K
    return max_sum;
}

static int lower_bound(ArrayList<Integer> al,
                       int x) 
{ 
    
    // x is the target value or key
    int l = -1, r = al.size();
    while (l + 1 < r)
    {
        int m = (l + r) >>> 1;
        if (al.get(m) >= x) 
            r = m;
        else
            l = m;
    }
    return r;
}

// Driver code
public static void main(String args[])
{
 
    // Initialise the array
    int arr[] = { 5, -2, 6, 3, -5 };
 
    // Initialise the value of K
    int K = 15;
 
    // Size of array
    int N = arr.length;
 
    System.out.println(maxSubarraySum(arr, N, K));
}
}

// This code is contributed by jyoti369

Python3

import bisect

# Function to maximum required sum < K
def maxSubarraySum(arr, N, K):
    # Hash to lookup for value (cum_sum - K)
    cum_set = set()
    cum_set.add(0)
    max_sum = float('-inf')
    cSum = 0
    for i in range(N):
        # getting cumulative sum from [0 to i]
        cSum += arr[i]

        # lookup for upperbound of (cSum-K) in hash
        al = [x for x in cum_set]
        al.sort()
        lower_bound_index = bisect.bisect_left(al, cSum - K)

        # check if upper_bound of (cSum-K) exists then update max sum
        if lower_bound_index != len(al):
            max_sum = max(max_sum, cSum - al[lower_bound_index])

        # // insert cumulative value in hash
        cum_set.add(cSum)

    #  return maximum sum lesser than K
    return max_sum


arr = [5, -2, 6, 3, -5]
K = 15
N = len(arr)
print(maxSubarraySum(arr, N, K))

// Java program to find maximum sum
// subarray less than K
using System;
using System.Collections.Generic;
class GFG {

    // Function to maximum required sum < K
    static int maxSubarraySum(int[] arr, int N, int K)
    {

        // Hash to lookup for value (cum_sum - K)
        HashSet<int> cum_set = new HashSet<int>();
        cum_set.Add(0);
        int max_sum = Int32.MinValue, cSum = 0;
        for (int i = 0; i < N; i++) {

            // Getting cumulative sum from [0 to i]
            cSum += arr[i];

            // Lookup for upperbound
            // of (cSum-K) in hash
            List<int> al = new List<int>();
            int end = 0;
            foreach(int it in cum_set)
            {
                end = it;
                al.Add(it);
            }

            al.Sort();
            int sit = lower_bound(al, cSum - K);

            // Check if upper_bound
            // of (cSum-K) exists
            // then update max sum
            if (sit != end)
                max_sum = Math.Max(max_sum, cSum - al.ElementAt(sit));

            // Insert cumulative value in hash
            cum_set.Add(cSum);
        }

        // Return maximum sum
        // lesser than K
        return max_sum;
    }
    static int lower_bound(List<int> al, int x)
    {

        // x is the target value or key
        int l = -1, r = al.Count;
        while (l + 1 < r) {
            int m = (l + r) >> 1;
            if (al[m] >= x)
                r = m;
            else
                l = m;
        }
        return r;
    }

    // Driver code
    public static void Main(string[] args)
    {

        // Initialise the array
        int[] arr = { 5, -2, 6, 3, -5 };

        // Initialise the value of K
        int K = 15;

        // Size of array
        int N = arr.Length;
        Console.Write(maxSubarraySum(arr, N, K));
    }
}

// This code is contributed by chitranayal.

JavaScript

<script>

// JavaScript program to find maximum sum
// subarray less than K


    // Function to maximum required sum < K
    function maxSubarraySum(arr, N, K)
    {

        // Hash to lookup for value (cum_sum - K)
        let cum_set = new Set();

        cum_set.add(0);

        let max_sum = Number.MIN_SAFE_INTEGER;
        let cSum = 0;

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

            // Getting cumulative sum from [0 to i]
            cSum += arr[i];

            // Lookup for upperbound
            // of (cSum-K) in hash
            let al = [];
            let end = 0;
            for(let it of cum_set)
            {
                end = it;
                al.push(it);
            }

            al.sort((a, b) => a - b);
            let sit = lower_bound(al, cSum - K);

            // Check if upper_bound
            // of (cSum-K) exists
            // then update max sum
            if (sit != end)
                max_sum = Math.max(max_sum, cSum - sit);

            // Insert cumulative value in hash
            cum_set.add(cSum);
        }

        // Return maximum sum
        // lesser than K
        return max_sum;
    }
    
    let lower_bound = 
    (al, x) => al.filter((item) => item > x )[0]     

    // Driver code


        // Initialise the array
        let arr = [ 5, -2, 6, 3, -5 ];

        // Initialise the value of K
        let K = 15;

        // Size of array
        let N = arr.length;
        document.write(maxSubarraySum(arr, N, K));

// This code is contributed by _saurabh_jaiswal

</script>

Output

Time Complexity: O(N*Log(N)), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.

Sanjit_Prasad

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