Pattern: Sliding Window — Problem Statement & Solution Guide
Problem Description
You are given an array of integers, nums, and an integer k. Your task is to determine the largest possible sum of any contiguous subarray whose length is exactly k. If the array contains fewer than k elements, the answer should be 0. The solution must run in linear time by maintaining a sliding window of size k and updating its sum as the window moves across the array, thereby avoiding recomputation of the sum from scratch for each position.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pattern: Sliding Window"
WHY DOES IT MATTER?
Sliding windows enable constant‑time updates for contiguous segments, turning potentially quadratic problems into linear ones. This pattern is essential for real‑time analytics, streaming data processing, and any scenario where data arrives continuously and decisions must be made on the fly.
OPTIMIZATION CHALLENGE
The key insight is that the sum of the next window differs from the current window by only two elements: one leaving and one entering. Recognizing this allows us to avoid recomputing the entire sum, reducing time from O(nk) to O(n) and space from O(k) to O(1).
REAL-WORLD CONNECTION
Consider a network router that monitors the last k packets to detect congestion. Instead of re‑calculating the total size of those packets each time a new packet arrives, the router subtracts the size of the packet that leaves the window and adds the new packet’s size—exactly the sliding window update.
When explaining the solution, emphasize the invariant: "currentSum always equals the sum of the last k elements processed." This helps interviewers see that the algorithm maintains correctness while achieving optimal performance.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The sliding window technique is a classic linear‑time strategy for problems that involve contiguous subarrays or substrings of a fixed size. In the naive approach, one would compute the sum of every possible subarray of length k by iterating over each starting index and summing k elements, leading to an O(nk) time complexity that quickly becomes infeasible for large arrays. The sliding window eliminates this redundancy by maintaining the sum of the current window and updating it in constant time as the window moves one step to the right: subtract the element that leaves the window and add the new element that enters. This transforms the algorithm into O(n) time while using only O(1) additional space, making it ideal for real‑time analytics, streaming data, and large‑scale systems where performance is critical.
The underlying theory hinges on the observation that the sum of a window of size k ending at index i can be derived from the sum of the previous window ending at i‑1 by a simple arithmetic operation. This incremental update property is what distinguishes sliding windows from other techniques like prefix sums or divide‑and‑conquer, which may still require O(n) preprocessing or O(log n) queries but not the same constant‑time update per step. By leveraging this property, we avoid recomputing sums from scratch and achieve optimal linear performance.
In practice, sliding windows are ubiquitous: from computing moving averages in finance, to detecting anomalies in sensor streams, to optimizing cache eviction policies. Understanding this pattern equips engineers to solve a wide range of problems efficiently and to explain their solutions clearly in interviews.
Interview Questions on This Problem
Q1How would you modify the sliding window solution if the subarray length k were variable and you needed the maximum sum over all lengths up to k?
You would maintain two pointers and a running sum, expanding the right pointer while the window size is less than or equal to k, and updating the maximum whenever the window size reaches k. If you need to consider all lengths up to k, you can keep a running maximum of sums for each window size by storing intermediate sums or by using a deque to track maximums for varying lengths, but the core idea remains incremental updates and careful pointer management.
Q2During a recent interview at a fintech company, the interviewer asked: "What is the time complexity if we use a prefix sum array instead of a sliding window?"
Using a prefix sum array, you can compute any subarray sum in O(1) after O(n) preprocessing. However, to find the maximum sum over all windows of size k, you would still need to iterate over all n−k+1 windows, resulting in O(n) time overall. The space complexity would be O(n) for the prefix array, which is higher than the O(1) space of the sliding window.
Q3A high‑growth startup asked: "Can you explain why the sliding window approach is preferable over a divide‑and‑conquer method for this problem?"
Divide‑and‑conquer would split the array recursively and combine results, typically leading to O(n log n) time for this specific problem. Sliding window processes each element exactly once with constant‑time updates, achieving O(n) time and O(1) space, which is both faster and more memory‑efficient—critical for large data streams in a startup environment.
Examples
Input
nums = [1, 2, 3, 4, 5], k = 3
Output
12
Explanation: The possible windows of length 3 are [1,2,3] (sum 6), [2,3,4] (sum 9), and [3,4,5] (sum 12). The maximum sum is 12.
Input
nums = [5, -1, 3, 4], k = 2
Output
7
Explanation: Windows: [5,-1] sum 4, [-1,3] sum 2, [3,4] sum 7. The maximum is 7.
Input
nums = [10, -2, 3, 1, 0], k = 4
Output
12
Explanation: Windows: [10,-2,3,1] sum 12, [-2,3,1,0] sum 2. The maximum is 12.
Input
nums = [1, 2], k = 5
Output
0
Explanation: The array length is less than k, so the result is 0.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- 1 <= k <= 100000
- k <= 100000
Optimal Approach & Strategy
Use a sliding window: keep a running sum of the current k elements, update it by subtracting the element that exits and adding the new one, and track the maximum. This runs in O(n) time and O(1) space.
Brute Force Approach
Compute the sum of every subarray of length k by nested loops: for each starting index, sum k elements. This takes O(nk) time and O(1) space.
Verified Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maxSumSubarray = function(nums, k) {
const n = nums.length;
if (n < k) return 0;
let currentSum = 0;
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let maxSum = currentSum;
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
};class Solution {
public:
int maxSumSubarray(vector<int>& nums, int k) {
int n = nums.size();
if (n < k) return 0;
int currentSum = 0;
for (int i = 0; i < k; i++) {
currentSum += nums[i];
}
int maxSum = currentSum;
for (int i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int maxSumSubarray(int[] nums, int k) {
int n = nums.length;
if (n < k) return 0;
int currentSum = 0;
for (int i = 0; i < k; i++) {
currentSum += nums[i];
}
int maxSum = currentSum;
for (int i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}class Solution:
def maxSumSubarray(self, nums: List[int], k: int) -> int:
n = len(nums)
if n < k:
return 0
current_sum = sum(nums[:k])
max_sum = current_sum
for i in range(k, n):
current_sum += nums[i] - nums[i - k]
max_sum = max(max_sum, current_sum)
return max_sum/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var maxSumSubarray = function(nums, k) {
const n = nums.length;
if (n < k) return 0;
let currentSum = 0;
for (let i = 0; i < k; i++) {
currentSum += nums[i];
}
let maxSum = currentSum;
for (let i = k; i < n; i++) {
currentSum += nums[i] - nums[i - k];
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
};Asked in Top Tech Interviews
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