239. Sliding Window Maximum

Given an array nums, 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 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       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7

Therefore, return the max sliding window as [3,3,5,5,6,7].

Note: 
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

Similar:

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155. Min Stack

159. Longest Substring with At Most Two Distinct Characters

265. Paint House II

Solution 1. Priority Queue
public class Solution {
    class Pair implements Comparable<Pair>{
        public int key;
        public int idx;
        
        public Pair(int k, int id) {
            this.key = k;
            this.idx = id;
        }
        
        @Override
        public int compareTo(Pair p) {
            return this.key < p.key ? 1 : (this.key == p.key ? 0 : -1); // reverse order, from Large to Small
        }
    }
    
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums.length == 0 || k <= 0) return new int[0];
        
        PriorityQueue<Pair> que = new PriorityQueue<Pair>();
        int[] ret = new int[nums.length - k + 1];
        int i = 0;
        
        for (i = 0; i < nums.length; i++) {
            que.add(new Pair(nums[i], i));
            while (que.peek().idx < i - k + 1) {
                que.poll();
            }
            if (i >= k-1) {
                ret[i-k+1] = que.peek().key;
            }
        }
        
        return ret;
    }
}