📏

Fixed Sliding Window

Window of exactly k elements slides one position at a time. Build once in O(k), then add right element − remove left element in O(1) per step.

Array String Time O(n) Space O(1)

📋 When to Use

Window size k is fixed and given in the problem
Find max / min / average of every window of size k
Count elements matching a condition in each window
Subarray / substring of length exactly k

💡 Naive O(n·k) recomputes each window. Sliding window maintains a running total for O(n) — k× faster.

⚡ Complexity

Time

O(n)

Space

O(1)

Each element is touched exactly twice — once entering, once leaving. Single pass, constant extra memory (just a few integers).

Problem Given an integer array nums and an integer k, find the maximum average value of any contiguous subarray of length k. Here we show maximum sum for clarity — divide by k to get the average.
Input: nums = [2, 1, 5, 1, 3, 2], k = 3 · Answer: window [5,1,3] → sum = 9, avg = 3.0

Step 1 / —

🔨 Phase 1: Build (i = 0 → k−1)

🪟 Phase 2: Slide (i = k → n−1)

✅ Done

nums = [2, 1, 5, 1, 3, 2] k = 3

windowSum = —

maxSum = —

window = —

Click ▶ Play or Next ▶ to start

Initialize: int windowSum = 0, maxSum = 0;

Complete Java TemplateJAVA

int maxWindowSum(int[] nums, int k) {
    int windowSum = 0;

    // ── Phase 1: Build first window ──────────────────────
    for (int i = 0; i < k; i++) {
        windowSum += nums[i];
    }
    int maxSum = windowSum;

    // ── Phase 2: Slide window one step at a time ─────────
    for (int i = k; i < nums.length; i++) {
        windowSum += nums[i] - nums[i - k];   // add right, remove left
        maxSum = Math.max(maxSum, windowSum);
    }

    return maxSum;   // O(n) time, O(1) space
}

🎯 Practice Problems

#643Maximum Average Subarray IEasy #219Contains Duplicate II — fixed window + HashSetEasy #1343Sub-arrays of Size K and Average ≥ ThresholdMedium #2461Maximum Sum of Distinct Subarrays With Length KMedium #239Sliding Window Maximum — fixed window + monotonic dequeHard