239. Sliding Window Maximum

You are given an array of integers 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.

Return the max sliding window.

Example 1:

Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation: 
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

Example 2:

Input: nums = [1], k = 1
Output: [1]

Solution:

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        // Assumptions: array is not null or not empty,
        // k >= 1 and k <= a.length.
        List<Integer> max = new ArrayList<Integer>();
        // use a descending deque to solve this problem,
        // we store the index instead of the actual value in the deque,
        // and we make sure:
        // 1. the deque only contains index in the current sliding window.
        // 2. for any index, the previous index with smaller value is 
        // discarded from the deque.
        Deque<Integer> deque = new LinkedList<Integer>();
        for (int i = 0; i < nums.length; i++){
            // discard any index with smaller value than index i.
            while(!deque.isEmpty() && nums[deque.peekLast()] <= nums[i]){
                deque.pollLast();
            }
            // it is possible the head element is out of the current
            // sliding window so we might need to discard it as well.
            if (!deque.isEmpty() && deque.peekFirst() <= i - k){
                deque.pollFirst();
            }
            deque.offerLast(i);
            if (i >= k - 1){
                max.add(nums[deque.peekFirst()]);
            }

        }
        int[] result = new int[max.size()];
        for (int i = 0; i < max.size(); i++){
            result[i] = max.get(i);
        }

        return result;

    }
}

// TC: O(n)
// SC: O(n)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int[] result = new int[nums.length - k + 1];
        // List<Integer> max = new ArrayList<Integer>();

        Deque<Integer> deque = new LinkedList<Integer>();

        for (int i = 0; i < nums.length; i++){
            while(!deque.isEmpty() && nums[deque.peekLast()] <= nums[i]){
                deque.pollLast();
            }


            // size
            if (!deque.isEmpty() && deque.peekFirst() <= i - k){ //     2 - 3 = -1
                deque.pollFirst();
            }


            deque.offerLast(i);
            if (i >= k - 1){//2>= 3 -1 =2  // make sure a sliding window size
                //  max.add(nums[deque.peekFirst()]);
                result[i -k + 1] = nums[deque.peekFirst()]; 
            }


        }


        /*
                     0 1  2  3 4 5 6 7
                                     i
                    [1,3,-1,-3,5,3,6,7], k = 3


        deque:   first   <-   7  ->   last
        max: 3, 3 , 5, 5,6 , 7
        */


        // int[] result = new int[max.size()];
        // for (int i = 0; i < max.size(); i++){
        //     result[i] = max.get(i);
        // }

        return result;

    }
}

// TC: O(n)
// SC: O(n)