Minimum Window Sum — Problem Statement & Solution Guide

Sliding WindowMediumSliding Window

TimeO(n)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Sliding Window and solve the Minimum Window Sum problem optimally.

TopicSliding Window

PatternSliding Window

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums and a positive integer target, determine the minimum possible length of a contiguous subarray whose elements sum to at least target. If no such subarray exists, return 0. The algorithm must run in O(n) time and O(1) additional space.

DSA Pattern Breakdown

"Minimum Window Sum"

medium

WHY DOES IT MATTER?

The sliding‑window pattern turns problems that ask for a contiguous sub‑structure with a numeric constraint into linear‑time solutions. It eliminates the need for nested loops, reduces cache misses, and is directly applicable to streaming scenarios where only O(1) extra memory is permissible.

OPTIMIZATION CHALLENGE

The key insight is that the sum of a window can be updated incrementally: add the new right‑most element when expanding and subtract the left‑most element when contracting. This constant‑time update means each element is processed at most twice, collapsing the quadratic search space into a single linear pass.

REAL-WORLD CONNECTION

Think of a network router that buffers packets until a certain byte threshold is reached before forwarding them. The router continuously adds incoming packets (expanding the window) and, once the threshold is met, starts dropping the oldest packets (contracting the window) to keep the buffer size minimal while still satisfying the bandwidth requirement.

During an interview, write the two‑pointer skeleton first, then immediately add the running sum variable. Test the shrink‑while‑valid loop early – it’s where the minimum length is captured. Remember to handle the "no solution" case by initializing the answer with Infinity and returning 0 if it never changes.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The Minimum Window Sum problem is a textbook case for the sliding‑window paradigm. The goal is to locate the shortest contiguous segment of an array whose elements add up to at least a given target. A naïve solution would enumerate every possible sub‑array, compute its sum, and keep the smallest length that meets the condition – an O(n²) time algorithm that quickly becomes infeasible for arrays with millions of elements. The failure of the brute‑force method stems from redundant recomputation: each time the window slides by one position, the sum of the previous window is discarded even though most of its elements are still relevant.

The optimal approach leverages the fact that all numbers are non‑negative (or that we can treat the problem as “at least target” without negative cancellation). By maintaining two pointers – a left boundary and a right boundary – we expand the window until the running sum reaches or exceeds the target, then contract from the left to try to shrink the window while preserving the sum constraint. This two‑pointer technique guarantees that each element is visited at most twice (once when the right pointer includes it, once when the left pointer excludes it), delivering a linear O(n) runtime with O(1) auxiliary space. The sliding window thus transforms a quadratic exploration into a single pass, which is why it is the optimal paradigm for this class of problems.

Interview Questions on This Problem

Q1How would you adapt the minimum window sum solution to handle arrays that may contain negative numbers?

With negative numbers the simple monotonic expansion‑contraction property breaks down, because adding a negative can reduce the sum and a previously optimal window might become viable again. A common adaptation is to use a prefix‑sum array combined with a balanced binary search tree (or deque) to query the smallest prefix index that satisfies prefix[j] - prefix[i] >= target, which runs in O(n log n) time. In an interview, you can discuss why the pure sliding window no longer works and propose the prefix‑sum + BST approach as a trade‑off.

Q2A fintech platform needs to detect the smallest time window where transaction volume exceeds a regulatory threshold. Which aspects of the Minimum Window Sum algorithm are directly applicable?

The problem maps one‑to‑one: transaction amounts become the array elements, the regulatory threshold is the target, and the time window corresponds to the sub‑array length. The sliding‑window technique provides an O(n) solution that can run in real‑time on streaming data, ensuring the platform can flag violations instantly without storing the entire transaction history.

Q3In a high‑growth startup, engineers often need to optimize API latency by finding the shortest burst of requests that saturates a server. How would you explain the relevance of the Minimum Window Sum pattern to a non‑technical stakeholder?

I would describe it as looking for the smallest consecutive group of requests that together push the server over a load limit. By sliding a window over the request stream and adjusting its size only when the load is too high, we can pinpoint the exact burst length that causes latency spikes, enabling targeted throttling or scaling decisions.

Examples

Example 1

Input

nums = [2,3,1,2,4,3], target = 7

Output

Explanation: Starting with the leftmost element, expand the window until the sum reaches 8 (indices 0‑3). Then contract from the left: removing the first element drops the sum to 6, so the window is shifted right. Later the window covering indices 4‑5 ([4,3]) has sum 7 and length 2, which is the smallest achievable.

Example 2

Input

nums = [1,4,4], target = 4

Output

Explanation: The element at index 1 equals the target, so a single‑element window satisfies the condition; no shorter window exists.

Example 3

Input

nums = [1,1,1,1,1,1,1], target = 11

Output

Explanation: Even the sum of the entire array is 7, which is less than the target, therefore no contiguous subarray meets the requirement.

Constraints

1 <= nums.length <= 100000
1 <= target <= 1000000000
1 <= nums[i] <= 100000

Optimal Approach & Strategy

Use two pointers to maintain a sliding window and a running sum; expand the right pointer until the sum ≥ target, then move the left pointer inward to shrink the window while updating the minimum length. Each element is added and removed at most once, yielding O(n) time and O(1) space.

Brute Force Approach

Enumerate every possible start index, then for each start compute the cumulative sum until the target is reached or the array ends, tracking the smallest length that satisfies the condition. This double loop results in O(n²) time and quickly times out on large inputs.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number} target
 * @param {number[]} nums
 * @return {number}
 */
var minSubArrayLen = function(target, nums) {
    const n = nums.length;
    if (n === 0) return 0;
    
    let minLen = Infinity;
    let sum = 0;
    let left = 0;
    
    for (let right = 0; right < n; right++) {
        sum += nums[right];
        
        while (sum >= target) {
            minLen = Math.min(minLen, right - left + 1);
            sum -= nums[left];
            left++;
        }
    }
    
    return minLen === Infinity ? 0 : minLen;
};

// Test cases
console.log(minSubArrayLen(7, [2, 3, 1, 2, 4, 3])); // Expected: 2
console.log(minSubArrayLen(4, [1, 4, 4])); // Expected: 1
console.log(minSubArrayLen(11, [1, 1, 1, 1, 1])); // Expected: 0

#include <iostream>
#include <vector>
#include <climits>

using namespace std;

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        
        int minLen = INT_MAX;
        int sum = 0;
        int left = 0;
        
        for (int right = 0; right < n; right++) {
            sum += nums[right];
            
            while (sum >= target) {
                minLen = min(minLen, right - left + 1);
                sum -= nums[left];
                left++;
            }
        }
        
        return minLen == INT_MAX ? 0 : minLen;
    }
};

int main() {
    vector<int> nums1 = {2, 3, 1, 2, 4, 3};
    int target1 = 7;
    Solution sol;
    cout << sol.minSubArrayLen(target1, nums1) << endl; // Expected: 2
    
    vector<int> nums2 = {1, 4, 4};
    int target2 = 4;
    cout << sol.minSubArrayLen(target2, nums2) << endl; // Expected: 1
    
    vector<int> nums3 = {1, 1, 1, 1, 1};
    int target3 = 11;
    cout << sol.minSubArrayLen(target3, nums3) << endl; // Expected: 0
    
    return 0;
}

import java.util.*;

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        
        int minLen = Integer.MAX_VALUE;
        int sum = 0;
        int left = 0;
        
        for (int right = 0; right < n; right++) {
            sum += nums[right];
            
            while (sum >= target) {
                minLen = Math.min(minLen, right - left + 1);
                sum -= nums[left];
                left++;
            }
        }
        
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }
    
    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.minSubArrayLen(7, new int[]{2, 3, 1, 2, 4, 3})); // Expected: 2
        System.out.println(sol.minSubArrayLen(4, new int[]{1, 4, 4})); // Expected: 1
        System.out.println(sol.minSubArrayLen(11, new int[]{1, 1, 1, 1, 1})); // Expected: 0
    }
}

from typing import List

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        if n == 0:
            return 0
        
        min_len = float('inf')
        total_sum = 0
        left = 0
        
        for right in range(n):
            total_sum += nums[right]
            
            while total_sum >= target:
                min_len = min(min_len, right - left + 1)
                total_sum -= nums[left]
                left += 1
        
        return 0 if min_len == float('inf') else min_len

# Test cases
if __name__ == "__main__":
    sol = Solution()
    print(sol.minSubArrayLen(7, [2, 3, 1, 2, 4, 3]))  # Expected: 2
    print(sol.minSubArrayLen(4, [1, 4, 4]))  # Expected: 1
    print(sol.minSubArrayLen(11, [1, 1, 1, 1, 1]))  # Expected: 0

/**
 * @param {number} target
 * @param {number[]} nums
 * @return {number}
 */
var minSubArrayLen = function(target, nums) {
    const n = nums.length;
    if (n === 0) return 0;
    
    let minLen = Infinity;
    let sum = 0;
    let left = 0;
    
    for (let right = 0; right < n; right++) {
        sum += nums[right];
        
        while (sum >= target) {
            minLen = Math.min(minLen, right - left + 1);
            sum -= nums[left];
            left++;
        }
    }
    
    return minLen === Infinity ? 0 : minLen;
};

// Test cases
console.log(minSubArrayLen(7, [2, 3, 1, 2, 4, 3])); // Expected: 2
console.log(minSubArrayLen(4, [1, 4, 4])); // Expected: 1
console.log(minSubArrayLen(11, [1, 1, 1, 1, 1])); // Expected: 0

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