LeetCode 480: Sliding Window Median Solution

Master LeetCode problem 480 (Sliding Window Median), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

480. Sliding Window Median

Hard

Array Hash Table Sliding Window Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we can use two heaps - a max heap to store the left half of the window elements and a min heap to store the right half of the window elements. By maintaining these two heaps, we can efficiently find the median of the current window.

Here is the step-by-step algorithm:

Initialize two heaps: a max heap left to store the left half of the window elements and a min heap right to store the right half of the window elements.
Iterate over the input array using a sliding window of size k:
- Add elements to the heaps based on their values.
- Balance the heaps if necessary by transferring elements between them.
- Calculate the median based on the top elements of the heaps.
- Add the median to the result array.
- Remove the element that goes out of the sliding window.
Return the result array containing the medians of all sliding windows.

Time Complexity: O(n log k) - We iterate through all elements in the input array and perform log k operations for heap balancing in each iteration. Space Complexity: O(k) - The space used by the two heaps.

Solution Code

import java.util.*;

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        PriorityQueue<Integer> left = new PriorityQueue<>(Collections.reverseOrder());
        PriorityQueue<Integer> right = new PriorityQueue<>();
        List<Double> medians = new ArrayList<>();

        for (int i = 0; i < nums.length; i++) {
            if (left.size() <= right.size()) {
                right.offer(nums[i]);
                left.offer(right.poll());
            } else {
                left.offer(nums[i]);
                right.offer(left.poll());
            }

            if (left.size() + right.size() == k) {
                double median;
                if (left.size() == right.size()) {
                    median = ((double) left.peek() + right.peek()) / 2.0;
                } else {
                    median = (double) left.peek();
                }
                medians.add(median);

                if (!left.remove(nums[i - k + 1])) {
                    right.remove(nums[i - k + 1]);
                }
            }
        }

        return medians.stream().mapToDouble(Double::doubleValue).toArray();
    }
}

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