Sliding window maximum

We can solve this problem using monotonic increasing queue. We will add all the items from the input lust in a monotonic q and add the top of the q to our result. Then when the traverse is finished, we slice top k elements from our result and return it.

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        q, res = collections.deque(), []

        for i in range(len(nums)):
            while q and q[0] <= i-k:
                q.popleft()
            while q and nums[i] >= nums[q[-1]]:
                q.pop()
            q.append(i)

            res.append(nums[q[0]])

        return res[k-1:]

Time Complexity: O(n)
Space Complexity: O(n)