Smallest sum contiguous subarray

Last Updated : 07 Mar, 2025

Given an array containing n integers. The problem is to find the sum of the elements of the contiguous subarray having the smallest(minimum) sum.

Examples:

Input: arr[] = {3, -4, 2, -3, -1, 7, -5}
Output: -6
Explanation: Subarray with minimum sum is {-4, 2, -3, -1} = -6
Input: arr = {2, 6, 8, 1, 4}
Output: 1
Explanation: Subarray with minimum sum is {1} = 1

[Naive Approach] - Using Nested Loops to Find all Subarray and their Sum - O(n^2) Time and O(1) Space

Consider all the contiguous subarrays of different sizes and find their sum. The subarray having the smallest(minimum) sum is the required answer.

C++

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

int smallestSumSubarr(vector<int> arr)
{
    int n = arr.size();
    int min_sum = INT_MAX; 
    int temp = 0;
    for( int i=0; i<n; i++)
    {
        temp =0; 
        for( int j=i; j<n; j++)
        {
            temp+= arr[j];
            min_sum = min(min_sum, temp);
        }
    }
    return min_sum; 
}


int main()
{
    vector<int> arr = {3, -4, 2, -3, -1, 7, -5};
    cout <<smallestSumSubarr(arr);
    return 0;
}

Java

import java.util.Arrays;
import java.util.List;

public class Main {
    public static int smallestSumSubarr(List<Integer> arr) {
        int n = arr.size();
        int min_sum = Integer.MAX_VALUE;
        int temp;
        for (int i = 0; i < n; i++) {
            temp = 0;
            for (int j = i; j < n; j++) {
                temp += arr.get(j);
                min_sum = Math.min(min_sum, temp);
            }
        }
        return min_sum;
    }

    public static void main(String[] args) {
        List<Integer> arr = Arrays.asList(3, -4, 2, -3, -1, 7, -5);
        System.out.println(smallestSumSubarr(arr));
    }
}

Python

def smallest_sum_subarr(arr):
    n = len(arr)
    min_sum = float('inf')
    for i in range(n):
        temp = 0
        for j in range(i, n):
            temp += arr[j]
            min_sum = min(min_sum, temp)
    return min_sum

arr = [3, -4, 2, -3, -1, 7, -5]
print(smallest_sum_subarr(arr))

using System;
using System.Collections.Generic;

class Program {
    static int SmallestSumSubarr(List<int> arr) {
        int n = arr.Count;
        int min_sum = int.MaxValue;
        int temp;
        for (int i = 0; i < n; i++) {
            temp = 0;
            for (int j = i; j < n; j++) {
                temp += arr[j];
                min_sum = Math.Min(min_sum, temp);
            }
        }
        return min_sum;
    }

    static void Main() {
        List<int> arr = new List<int> { 3, -4, 2, -3, -1, 7, -5 };
        Console.WriteLine(SmallestSumSubarr(arr));
    }
}

JavaScript

function smallestSumSubarr(arr) {
    let n = arr.length;
    let min_sum = Infinity;
    for (let i = 0; i < n; i++) {
        let temp = 0;
        for (let j = i; j < n; j++) {
            temp += arr[j];
            min_sum = Math.min(min_sum, temp);
        }
    }
    return min_sum;
}

const arr = [3, -4, 2, -3, -1, 7, -5];
console.log(smallestSumSubarr(arr));

Output

-6

[Expected Approach] - Using Kadane's Algirithm - O(n) Time and O(1) Space

It is a variation to the problem of finding the largest sum contiguous subarray based on the idea of Kadane’s algorithm.
The idea is to traverse over the array from left to right and for each element, find the minimum sum sum among all subarrays ending at that element. The result will be the maximum of all these values.
But, the main issue is how to calculate minimum sum among all the subarrays ending at an element in O(n) time?
To calculate the minimum sum of subarray ending at current element, say min_ending_here, we can use the minimum sum ending at the previous element. So for any element, we have two choices:
Choice 1: Extend the minimum sum subarray ending at the previous element by adding the current element to it. If the minimum subarray sum ending at the previous index is negative, then it is always better to extend the subarray.
Choice 2: Start a new subarray starting from the current element. If the minimum subarray sum ending at the previous index is positive, it is always better to start a new subarray from the current element.

C++

#include <bits/stdc++.h>

using namespace std;

// function to find the smallest sum contiguous subarray
int smallestSumSubarr(int arr[], int n)
{
    // to store the minimum value that is ending
    // up to the current index
    int min_ending_here = INT_MAX;

    // to store the minimum value encountered so far
    int min_so_far = INT_MAX;

    // traverse the array elements
    for (int i = 0; i < n; i++)
    {
        // if min_ending_here > 0, then it could not possibly
        // contribute to the minimum sum further
        if (min_ending_here > 0)
            min_ending_here = arr[i];

        // else add the value arr[i] to min_ending_here
        else
            min_ending_here += arr[i];

        // update min_so_far
        min_so_far = min(min_so_far, min_ending_here);
    }

    // required smallest sum contiguous subarray value
    return min_so_far;
}

int main()
{
    int arr[] = {3, -4, 2, -3, -1, 7, -5};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout <<smallestSumSubarr(arr, n);
    return 0;
}

Java

import java.io.*;
class GFG {

    // function to find the smallest sum contiguous
    // subarray
    static int smallestSumSubarr(int arr[], int n)
    {

        // to store the minimum value that is
        // ending up to the current index
        int min_ending_here = 2147483647;

        // to store the minimum value encountered
        // so far
        int min_so_far = 2147483647;

        // traverse the array elements
        for (int i = 0; i < n; i++) {

            // if min_ending_here > 0, then it could
            // not possibly contribute to the
            // minimum sum further
            if (min_ending_here > 0)
                min_ending_here = arr[i];

            // else add the value arr[i] to
            // min_ending_here
            else
                min_ending_here += arr[i];

            // update min_so_far
            min_so_far
                = Math.min(min_so_far, min_ending_here);
        }

        // required smallest sum contiguous
        // subarray value
        return min_so_far;
    }

    public static void main(String[] args)
    {

        int arr[] = { 3, -4, 2, -3, -1, 7, -5 };
        int n = arr.length;

        System.out.print(smallestSumSubarr(arr, n));
    }
}

// This code is contributed by Anant Agarwal.

Python

maxsize = int(1e9 + 7)


def smallestSumSubarr(arr, n):
    # to store the minimum value that is ending
    # up to the current index
    min_ending_here = maxsize

    # to store the minimum value encountered so far
    min_so_far = maxsize

    # traverse the array elements
    for i in range(n):
        # if min_ending_here > 0, then it could not possibly
        # contribute to the minimum sum further
        if (min_ending_here > 0):
            min_ending_here = arr[i]

        # else add the value arr[i] to min_ending_here
        else:
            min_ending_here += arr[i]

        # update min_so_far
        min_so_far = min(min_so_far, min_ending_here)

    # required smallest sum contiguous subarray value
    return min_so_far


arr = [3, -4, 2, -3, -1, 7, -5]
n = len(arr)
print(smallestSumSubarr(arr, n))

using System;

class GFG {
    // function to find the smallest sum
    // contiguous subarray
    static int smallestSumSubarr(int[] arr, int n)
    {
        // to store the minimum value that is
        // ending up to the current index
        int min_ending_here = 2147483647;
        // to store the minimum value encountered
        // so far
        int min_so_far = 2147483647;
        // traverse the array elements
        for (int i = 0; i < n; i++) {

            // if min_ending_here > 0, then it could
            // not possibly contribute to the
            // minimum sum further
            if (min_ending_here > 0)
                min_ending_here = arr[i];

            // else add the value arr[i] to
            // min_ending_here
            else
                min_ending_here += arr[i];

            // update min_so_far
            min_so_far
                = Math.Min(min_so_far, min_ending_here);
        }
        // required smallest sum contiguous
        // subarray value
        return min_so_far;
    }

    public static void Main()
    {
        int[] arr = { 3, -4, 2, -3, -1, 7, -5 };
        int n = arr.Length;

        Console.Write(smallestSumSubarr(arr, n));
    }
}

JavaScript

function smallestSumSubarr(arr, n)
{
    // to store the minimum value that is
    // ending up to the current index
    let min_ending_here = 2147483647;

    // to store the minimum value encountered
    // so far
    let min_so_far = 2147483647;

    // traverse the array elements
    for (let i = 0; i < n; i++) {

        // if min_ending_here > 0, then it could
        // not possibly contribute to the
        // minimum sum further
        if (min_ending_here > 0)
            min_ending_here = arr[i];

        // else add the value arr[i] to
        // min_ending_here
        else
            min_ending_here += arr[i];

        // update min_so_far
        min_so_far = Math.min(min_so_far, min_ending_here);
    }

    // required smallest sum contiguous
    // subarray value
    return min_so_far;
}

let arr = [ 3, -4, 2, -3, -1, 7, -5 ];
let n = arr.length;

console.log(smallestSumSubarr(arr, n));

Output

-6

Ayush

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Article Tags :

Smallest sum contiguous subarray

[Naive Approach] - Using Nested Loops to Find all Subarray and their Sum - O(n^2) Time and O(1) Space

[Expected Approach] - Using Kadane's Algirithm - O(n) Time and O(1) Space

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