LeetCode #480 — HARD

Sliding Window Median

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.

  • For examples, if arr = [2,3,4], the median is 3.
  • For examples, if arr = [1,2,3,4], the median is (2 + 3) / 2 = 2.5.

You are given an integer array nums and an integer k. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.

Return the median array for each window in the original array. Answers within 10-5 of the actual value will be accepted.

Example 1:

Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
Explanation: 
Window position                Median
---------------                -----
[1  3  -1] -3  5  3  6  7        1
 1 [3  -1  -3] 5  3  6  7       -1
 1  3 [-1  -3  5] 3  6  7       -1
 1  3  -1 [-3  5  3] 6  7        3
 1  3  -1  -3 [5  3  6] 7        5
 1  3  -1  -3  5 [3  6  7]       6

Example 2:

Input: nums = [1,2,3,4,2,3,1,4,2], k = 3
Output: [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]

Constraints:

  • 1 <= k <= nums.length <= 105
  • -231 <= nums[i] <= 231 - 1
Patterns Used
🪟Sliding Window→⚖Two Heaps→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values. For examples, if arr = [2,3,4], the median is 3. For examples, if arr = [1,2,3,4], the median is (2 + 3) / 2 = 2.5. You are given an integer array nums and an integer k. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the median array for each window in the original array. Answers within 10-5 of the actual value will be accepted.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Sliding Window

Example 1

[1,3,-1,-3,5,3,6,7]
3

Example 2

[1,2,3,4,2,3,1,4,2]
3

Related Problems

  • Find Median from Data Stream (find-median-from-data-stream)
  • Minimum Operations to Make Median of Array Equal to K (minimum-operations-to-make-median-of-array-equal-to-k)
Step 02

Core Insight

What unlocks the optimal approach

  • The simplest of solutions comes from the basic idea of finding the median given a set of numbers. We know that by definition, a median is the center element (or an average of the two center elements). Given an unsorted list of numbers, how do we find the median element? If you know the answer to this question, can we extend this idea to every sliding window that we come across in the array?
  • Is there a better way to do what we are doing in the above hint? Don't you think there is duplication of calculation being done there? Is there some sort of optimization that we can do to achieve the same result? This approach is merely a modification of the basic approach except that it simply reduces duplication of calculations once done.
  • The third line of thought is also based on this same idea but achieving the result in a different way. We obviously need the window to be sorted for us to be able to find the median. Is there a data-structure out there that we can use (in one or more quantities) to obtain the median element extremely fast, say O(1) time while having the ability to perform the other operations fairly efficiently as well?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #480: Sliding Window Median
class MedianFinder {
    private PriorityQueue<Integer> small = new PriorityQueue<>(Comparator.reverseOrder());
    private PriorityQueue<Integer> large = new PriorityQueue<>();
    private Map<Integer, Integer> delayed = new HashMap<>();
    private int smallSize;
    private int largeSize;
    private int k;

    public MedianFinder(int k) {
        this.k = k;
    }

    public void addNum(int num) {
        if (small.isEmpty() || num <= small.peek()) {
            small.offer(num);
            ++smallSize;
        } else {
            large.offer(num);
            ++largeSize;
        }
        rebalance();
    }

    public double findMedian() {
        return (k & 1) == 1 ? small.peek() : ((double) small.peek() + large.peek()) / 2;
    }

    public void removeNum(int num) {
        delayed.merge(num, 1, Integer::sum);
        if (num <= small.peek()) {
            --smallSize;
            if (num == small.peek()) {
                prune(small);
            }
        } else {
            --largeSize;
            if (num == large.peek()) {
                prune(large);
            }
        }
        rebalance();
    }

    private void prune(PriorityQueue<Integer> pq) {
        while (!pq.isEmpty() && delayed.containsKey(pq.peek())) {
            if (delayed.merge(pq.peek(), -1, Integer::sum) == 0) {
                delayed.remove(pq.peek());
            }
            pq.poll();
        }
    }

    private void rebalance() {
        if (smallSize > largeSize + 1) {
            large.offer(small.poll());
            --smallSize;
            ++largeSize;
            prune(small);
        } else if (smallSize < largeSize) {
            small.offer(large.poll());
            --largeSize;
            ++smallSize;
            prune(large);
        }
    }
}

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        MedianFinder finder = new MedianFinder(k);
        for (int i = 0; i < k; ++i) {
            finder.addNum(nums[i]);
        }
        int n = nums.length;
        double[] ans = new double[n - k + 1];
        ans[0] = finder.findMedian();
        for (int i = k; i < n; ++i) {
            finder.addNum(nums[i]);
            finder.removeNum(nums[i - k]);
            ans[i - k + 1] = finder.findMedian();
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n × k) time
O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW
O(n) time
O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.

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#219Contains Duplicate IIeasy→#594Longest Harmonious Subsequenceeasy→#632Smallest Range Covering Elements from K Listshard→#904Fruit Into Basketsmedium→#930Binary Subarrays With Summedium→