deque.java

/*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:
  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
*/

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

        int n = nums.length;
        int[] res = new int[n - k + 1];
        Deque<Integer> deque = new LinkedList<>(); //use of deque aka a double-ended queue instead of declaring a priority queue in every cycle

        for (int i = 0; i < n; i++) {
            // Remove indices that are out of the current window
            if (!deque.isEmpty() && deque.peekFirst() == i - k) {
                deque.pollFirst();
            }

            // Remove elements from the back of the deque that are smaller than the current element
            while (!deque.isEmpty() && nums[deque.peekLast()] <= nums[i]) {
                deque.pollLast();
            }

            // Add current element index at the back of the deque
            deque.offerLast(i);

            // The front of the deque contains the index of the maximum element for the current window
            if (i >= k - 1) {
                res[i - k + 1] = nums[deque.peekFirst()];
            }
        }

        return res;
    }
}