239 Sliding Window Maximum

239. Sliding Window Maximum

1. Question

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, Givennums=[1,3,-1,-3,5,3,6,7], andk= 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 assumekis always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Follow up: Could you solve it in linear time?

2. Implementation

(1) Heap with PriorityQueue

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return new int[0];
        }

        int n = nums.length;
        int[] res = new int[n - k + 1];
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());

        for (int i = 0; i < n; i++) {
            maxHeap.add(nums[i]);

            if (i >= k) {
                maxHeap.remove(nums[i - k]);
            }

            if (i - k + 1 >= 0) {
                res[i - k + 1] = maxHeap.peek();
            }
        }
        return res;
    }
}

(2) Heap with TreeMap

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return new int[0];
        }

        int n = nums.length;
        int[] res = new int[n - k + 1];

        TreeMap<Integer, Integer> map = new TreeMap();

        for (int i = 0; i < n; i++) {
            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);

            if (i >= k) {
                map.put(nums[i - k], map.get(nums[i - k]) - 1);
                if (map.get(nums[i - k]) == 0) {
                    map.remove(nums[i - k]);
                }
            }

            if (i - k + 1 >= 0) {
                res[i - k + 1] = map.lastKey();
            }
        }
        return res;
    }
}

(3) Deque

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return new int[0];
        }

        int n = nums.length;
        int[] res = new int[n - k + 1];

        ArrayDeque<Integer> deque = new ArrayDeque();
        int index = 0;

        for (int i = 0; i < n; i++) {
            // Remove index that is out of slinding window 
            while (!deque.isEmpty() && deque.peek() < (i - k + 1)) {
                deque.remove();
            }

            // Remove number which is smaller than the current number since they won't be the max
            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
                deque.removeLast();
            }

            deque.add(i);

            if (i - k + 1 >= 0) {
                res[index++] = nums[deque.peek()];
            }
        }
        return res;
    }
}

3. Time & Space Complexity

Heap with PriorityQueue: 时间复杂度O(n*k)，PriorityQueue.remove(Object)要O(k)的时间，而PriorityQueue.remove()需要O(logk)的时间, 空间复杂度O(n) Heap with TreeMap: 时间复杂度O(nlogk), 空间复杂度(n) TreeMap需要O(k)空间， res需要n - k + 1空间，加起来要O(n)

Deque: 时间复杂度O(n), 空间复杂度O(n)