Smallest subarray with sum greater than a given value

Last Updated : 30 Mar, 2026

Given an array arr[] of integers and a number x, the task is to find the smallest subarray with a sum strictly greater than x.

Examples:

Input: x = 51, arr[] = [1, 4, 45, 6, 0, 19]
Output: 3
Explanation: Minimum length subarray is [4, 45, 6]

Input: x = 100, arr[] = [1, 10, 5, 2, 7]
Output: 0
Explanation: No subarray exist

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Table of Content

[Naive Approach] Using Two Nested Loops - O(n^2) Time and O(1) Space

The idea is to use two nested loops. The outer loop picks a starting element, the inner loop considers all elements (on right side of current start) as ending element. Whenever sum of elements between current start and end becomes greater than x, update the result if current length is smaller than the smallest length so far. 

C++ 
#include <iostream>
using namespace std;

int smallestSubWithSum(int x, vector<int> &arr)
{
    int n = arr.size();
    int res = INT_MAX;

    // Pick every element as starting point
    for (int i = 0; i < n; i++)
    {
        int curr = 0;

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

            if (curr > x)
            {
                res = min(res, j - i + 1);
                break;
            }
        }
    }

    // Return 0 if answer does not exist
    if (res == INT_MAX)
        return 0;

    return res;
}

int main()
{
    vector<int> arr = {1, 4, 45, 6, 10, 19};
    int x = 51;

    cout << smallestSubWithSum(x, arr);

    return 0;
}
Java 
class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {
        int n = arr.length;
        int res = Integer.MAX_VALUE;

        // Pick every element as starting point
        for (int i = 0; i < n; i++) {
            int curr = 0;

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

                if (curr > x) {
                    res = Math.min(res, j - i + 1);
                    break;
                }
            }
        }

        // Return 0 if answer does exist.
        if (res == Integer.MAX_VALUE)
            return 0;
        return res;
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        System.out.println(smallestSubWithSum(x, arr));
    }
}
Python 
def smallestSubWithSum(x, arr):
    n = len(arr)
    res = float('inf')

    # Pick every element as starting point
    for i in range(n):
        curr = 0

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

            if curr > x:
                res = min(res, j - i + 1)
                break

    # Return 0 if answer does not exist.
    if res == float('inf'):
        return 0
    return res


if __name__ == "__main__":
    arr = [1, 4, 45, 6, 10, 19]
    x = 51

    print(smallestSubWithSum(x, arr))
C# 
using System;
class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {
        int n = arr.Length;
        int res = int.MaxValue;

        // Pick every element as starting point
        for (int i = 0; i < n; i++) {
            int curr = 0;

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

                if (curr > x) {
                    res = Math.Min(res, j - i + 1);
                    break;
                }
            }
        }

        // Return 0 if answer does not exist.
        if (res == int.MaxValue)
            return 0;
        return res;
    }

    static void Main(string[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        Console.WriteLine(smallestSubWithSum(x, arr));
    }
}
Javascript 
function smallestSubWithSum(x, arr)
{
    let n = arr.length;
    let res = Infinity;

    // Pick every element as starting point
    for (let i = 0; i < n; i++) {
        let curr = 0;

        for (let j = i; j < n; j++) {
            curr += arr[j];

            if (curr > x) {
                res = Math.min(res, j - i + 1);
                break;
            }
        }
    }

    // Return 0 if answer does not exist.
    if (res === Infinity)
        return 0;
    return res;
}

// driver code
let arr = [ 1, 4, 45, 6, 10, 19 ];
let x = 51;

console.log(smallestSubWithSum(x, arr));

Output
3

[Better Approach] - Prefix Sum and Binary Search - O(n Log n) Time and O(n) Space

The idea is to store the prefix sum in an array and then for every index i, perform binary search in the range [i+1, n] to find the minimum index such that preSum[j] > preSum[i] + x.

Below is the step by step of above approach:

  • Compute prefix sum in an array preSum[].
  • Iterate through preSum[] and find lower bound for x + preSum[i], here lower bound means index of first value greater than x + preSum[i].
  • If the lower bound is found and it's not equal to x i.e., the subarray sum is greater than the x, calculate the length of current subarray and update result if the current result is a smaller value.
C++ 
#include <iostream>
using namespace std;
int smallestSubWithSum(int x, vector<int> &arr)
{
    int n = arr.size();
    int res = INT_MAX;
    vector<int> preSum(n + 1, 0);

    // Compute the prefix sums
    for (int i = 1; i <= n; i++)
        preSum[i] = preSum[i - 1] + arr[i - 1];

    // Iterate through each starting index
    for (int i = 1; i <= n; i++)
    {
        int toFind = x + preSum[i - 1];
        auto bound = lower_bound(preSum.begin(), preSum.end(), toFind);

        if (bound != preSum.end() && *bound != toFind)
        {
            int len = bound - (preSum.begin() + i - 1);
            res = min(res, len);
        }
    }

    // If subarray does not exists
    if (res == INT_MAX)
        return 0;

    return res;
}

int main()
{

    vector<int> arr = {1, 4, 45, 6, 10, 19};
    int x = 51;

    cout << smallestSubWithSum(x, arr);

    return 0;
}
Java 
import java.util.*;
class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {
        int n = arr.length;

        int res = Integer.MAX_VALUE;
        int[] preSum = new int[n + 1];

        // Compute the prefix sums
        for (int i = 1; i <= n; i++) {
            preSum[i] = preSum[i - 1] + arr[i - 1];
        }

        // Iterate through each starting index
        for (int i = 1; i <= n; i++) {
            int toFind = x + preSum[i - 1] + 1;
            int bound = Arrays.binarySearch(preSum, toFind);
            if (bound < 0) {
                bound = -(bound + 1);
            }

            if (bound <= n) {
                int len = bound - (i - 1);
                res = Math.min(res, len);
            }
        }

        // If subarray does not exists
        if (res == Integer.MAX_VALUE)
            return 0;
        return res;
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        System.out.println(smallestSubWithSum(x, arr));
    }
}
Python 
from bisect import bisect_left


def smallestSubWithSum(x, arr):
    n = len(arr)

    res = float('inf')
    preSum = [0] * (n + 1)

    # Compute the prefix sums
    for i in range(1, n + 1):
        preSum[i] = preSum[i - 1] + arr[i - 1]

    # Iterate through each starting index
    for i in range(1, n + 1):
        toFind = x + preSum[i - 1] + 1
        bound = bisect_left(preSum, toFind)
        if bound <= n:
            len_sub = bound - (i - 1)
            res = min(res, len_sub)

    # If subarray does not exists
    if res == float('inf'):
        return 0

    return res


if __name__ == "__main__":
    arr = [1, 4, 45, 6, 10, 19]
    x = 51

    print(smallestSubWithSum(x, arr))
C# 
using System;
class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {
        int n = arr.Length;

        int res = int.MaxValue;
        int[] preSum = new int[n + 1];

        // Compute the prefix sums
        for (int i = 1; i <= n; i++) {
            preSum[i] = preSum[i - 1] + arr[i - 1];
        }

        // Iterate through each starting index
        for (int i = 1; i <= n; i++) {
            int toFind = x + preSum[i - 1] + 1;
            int bound = Array.BinarySearch(preSum, toFind);
            if (bound < 0) {
                bound = ~bound;
            }

            if (bound <= n) {
                int len = bound - (i - 1);
                res = Math.Min(res, len);
            }
        }

        // If subarray does not exists
        if (res == int.MaxValue)
            return 0;

        return res;
    }

    static void Main(string[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        Console.WriteLine(smallestSubWithSum(x, arr));
    }
}
JavaScript 
function smallestSubWithSum(x, arr)
{
    let n = arr.length;

    let res = Infinity;
    let preSum = new Array(n + 1).fill(0);

    // Compute the prefix sums
    for (let i = 1; i <= n; i++) {
        preSum[i] = preSum[i - 1] + arr[i - 1];
    }

    // Iterate through each starting index
    for (let i = 1; i <= n; i++) {
        let toFind = x + preSum[i - 1] + 1;
        let bound = preSum.findIndex(val => val >= toFind);

        if (bound !== -1) {
            let len = bound - (i - 1);
            res = Math.min(res, len);
        }
    }

    // If subarray does not exists
    if (res === Infinity)
        return 0;

    return res;
}

// driver code
let arr = [ 1, 4, 45, 6, 10, 19 ];
let x = 51;

console.log(smallestSubWithSum(x, arr));

Output
3

[Expected Approach] - Using Sliding Window - O(n) Time and O(1) Space

The idea is to maintain a sliding window, where we keep expanding the window by adding elements until the sum becomes greater than x, then we try to minimize this window by shrinking it from the start while maintaining the sum > x condition. This way, we explore all possible subarrays and keep track of the smallest valid length.

C++ 
#include <iostream>
using namespace std;
int smallestSubWithSum(int x, vector<int> &arr)
{

    int i = 0, j = 0;
    int sum = 0;
    int ans = INT_MAX;

    while (j < arr.size())
    {
        while (j < arr.size() && sum <= x)
        {
            sum += arr[j++];
        }
        if (j == arr.size() && sum <= x)
            break;

        // Minimize window from start while maintaining sum > x
        while (i < j && sum - arr[i] > x)
        {
            sum -= arr[i++];
        }

        ans = min(ans, j - i);

        // Remove current start element and shift window
        sum -= arr[i];
        i++;
    }
    if (ans == INT_MAX)
        return 0;
    return ans;
}

int main()
{
    vector<int> arr = {1, 4, 45, 6, 10, 19};
    int x = 51;

    cout << smallestSubWithSum(x, arr);

    return 0;
}
Java 
import java.util.*;

class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {
        int i = 0, j = 0;
        int sum = 0;
        int ans = Integer.MAX_VALUE;

        while (j < arr.length) {
            while (j < arr.length && sum <= x) {
                sum += arr[j++];
            }
            if (j == arr.length && sum <= x)
                break;

            // Minimize window from start while maintaining
            // sum > x
            while (i < j && sum - arr[i] > x) {
                sum -= arr[i++];
            }

            ans = Math.min(ans, j - i);

            // Remove current start element and shift window
            sum -= arr[i];
            i++;
        }
        if (ans == Integer.MAX_VALUE)
            return 0;
        return ans;
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        System.out.println(smallestSubWithSum(x, arr));
    }
}
Python 
def smallestSubWithSum(x, arr):

    i, j = 0, 0
    sum = 0
    ans = float('inf')

    while j < len(arr):
        while j < len(arr) and sum <= x:
            sum += arr[j]
            j += 1
        if j == len(arr) and sum <= x:
            break

        # Minimize window from start while maintaining sum > x
        while i < j and sum - arr[i] > x:
            sum -= arr[i]
            i += 1

        ans = min(ans, j - i)

        # Remove current start element and shift window
        sum -= arr[i]
        i += 1
    if ans == float('inf'):
        return 0
    return ans


if __name__ == "__main__":
    arr = [1, 4, 45, 6, 10, 19]
    x = 51

    print(smallestSubWithSum(x, arr))
C# 
using System;

class GfG {
    static int smallestSubWithSum(int x, int[] arr)
    {

        int i = 0, j = 0;
        int sum = 0;
        int ans = int.MaxValue;

        while (j < arr.Length) {
            while (j < arr.Length && sum <= x) {
                sum += arr[j++];
            }
            if (j == arr.Length && sum <= x)
                break;

            // Minimize window from start while maintaining
            // sum > x
            while (i < j && sum - arr[i] > x) {
                sum -= arr[i++];
            }

            ans = Math.Min(ans, j - i);

            // Remove current start element and shift window
            sum -= arr[i];
            i++;
        }
        if (ans == int.MaxValue)
            return 0;
        return ans;
    }

    static void Main(string[] args)
    {
        int[] arr = { 1, 4, 45, 6, 10, 19 };
        int x = 51;

        Console.WriteLine(smallestSubWithSum(x, arr));
    }
}
Javascript 
function smallestSubWithSum(x, arr)
{

    let i = 0, j = 0;
    let sum = 0;
    let ans = Infinity;

    while (j < arr.length) {
        while (j < arr.length && sum <= x) {
            sum += arr[j++];
        }
        if (j === arr.length && sum <= x)
            break;

        // Minimize window from start while maintaining sum
        // > x
        while (i < j && sum - arr[i] > x) {
            sum -= arr[i++];
        }

        ans = Math.min(ans, j - i);

        // Remove current start element and shift window
        sum -= arr[i];
        i++;
    }
    if (ans === Infinity)
        return 0;
    return ans;
}

// driver code
let arr = [ 1, 4, 45, 6, 10, 19 ];
let x = 51;

console.log(smallestSubWithSum(x, arr));

Output
3
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