Top Sliding Window Interview Questions & Solutions (2026)

1. Introduction to the Sliding Window Pattern

The Sliding Window technique is used to transform nested O(N²) brute-force subarray/substring checks into optimal O(N) linear-time algorithms.

+─────────────────────────────────────────────────────────────────────────────+
|                         SLIDING WINDOW MECHANICS                            |
+─────────────────────────────────────────────────────────────────────────────+

 Array: [ 2,  1,  5,  1,  3,  2 ],  Window Size K = 3

 Step 1: [ 2,  1,  5 ] 1,  3,  2  --> Sum = 8
 Step 2:   2, [ 1,  5,  1 ] 3,  2  --> Sum = 8 - 2 + 1 = 7  (O(1) update!)
 Step 3:   2,  1, [ 5,  1,  3 ] 2  --> Sum = 7 - 1 + 3 = 9  (Max = 9!)
 Step 4:   2,  1,  5, [ 1,  3,  2 ] --> Sum = 9 - 5 + 2 = 6

2. Core Interview Questions & Solutions

Q1. Maximum Sum Subarray of Size K (Fixed Window)

Problem Statement: Given an array of integers nums and an integer k, find the maximum sum of any contiguous subarray of size k.

cpp
#include <vector>
#include <numeric>
#include <algorithm>

int maxSubArrayLenK(const std::vector<int>& nums, int k) {
    if (nums.size() < k) return 0;
    
    int windowSum = 0;
    for (int i = 0; i < k; i++) windowSum += nums[i];
    
    int maxSum = windowSum;
    for (int i = k; i < nums.size(); i++) {
        windowSum += nums[i] - nums[i - k]; // Slide window in O(1)
        maxSum = std::max(maxSum, windowSum);
    }
    return maxSum;
}

• Time Complexity: O(N) | Space Complexity: O(1)

Q2. Longest Substring Without Repeating Characters (Dynamic Window)

Problem Statement: Given a string s, find the length of the longest substring without duplicate characters.

python
def lengthOfLongestSubstring(s: str) -> int:
    char_index_map = {}
    left = 0
    max_len = 0
    
    for right, char in enumerate(s):
        if char in char_index_map and char_index_map[char] >= left:
            left = char_index_map[char] + 1 # Jump left pointer
            
        char_index_map[char] = right
        max_len = max(max_len, right - left + 1)
        
    return max_len

• Time Complexity: O(N) | Space Complexity: O(min(N, Sigma))

Q3. Minimum Window Substring (Advanced Dynamic Window)

Problem Statement: Given two strings s and t, return the minimum window substring of s such that every character in t (including duplicates) is included in the window.

java
import java.util.*;

class Solution {
    public String minWindow(String s, String t) {
        if (s.length() < t.length()) return "";

        Map<Character, Integer> targetCounts = new HashMap<>();
        for (char c : t.toCharArray()) {
            targetCounts.put(c, targetCounts.getOrDefault(c, 0) + 1);
        }

        int required = targetCounts.size();
        int formed = 0;
        Map<Character, Integer> windowCounts = new HashMap<>();

        int left = 0, right = 0;
        int[] ans = {-1, 0, 0}; // length, left, right

        while (right < s.length()) {
            char c = s.charAt(right);
            windowCounts.put(c, windowCounts.getOrDefault(c, 0) + 1);

            if (targetCounts.containsKey(c) && windowCounts.get(c).intValue() == targetCounts.get(c).intValue()) {
                formed++;
            }

            // Try contracting window from left
            while (left <= right && formed == required) {
                c = s.charAt(left);
                if (ans[0] == -1 || (right - left + 1) < ans[0]) {
                    ans[0] = right - left + 1;
                    ans[1] = left;
                    ans[2] = right;
                }

                windowCounts.put(c, windowCounts.get(c) - 1);
                if (targetCounts.containsKey(c) && windowCounts.get(c) < targetCounts.get(c)) {
                    formed--;
                }
                left++;
            }
            right++;
        }

        return ans[0] == -1 ? "" : s.substring(ans[1], ans[2] + 1);
    }
}

• Time Complexity: O(S + T) | Space Complexity: O(S + T)

3. Sliding Window Pattern Summary

Practice all sliding window problems on DSAMaster's practice platform.