_239.java

package com.fishercoder.solutions;

import java.util.PriorityQueue;

/**
 * 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?

 Hint:

 How about using a data structure such as deque (double-ended queue)?
 The queue size need not be the same as the window’s size.
 Remove redundant elements and the queue should store only elements that need to be considered.
 */
public class _239 {

    public static class Solution1 {
        public int[] maxSlidingWindow(int[] nums, int k) {
            if (nums == null || nums.length == 0 || k == 0) {
                return new int[0];
            }
            PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> b - a);
            int[] res = new int[nums.length - k + 1];
            for (int i = 0; i < nums.length; i++) {
                if (i < k) {
                    heap.offer(nums[i]);
                    if (i == k - 1) {
                        res[0] = heap.peek();
                    }
                } else {
                    heap.remove(nums[i - k]);
                    heap.offer(nums[i]);
                    res[i - k + 1] = heap.peek();
                }
            }
            return res;
        }
    }
}