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239. Sliding Window Maximum

Array Queue Sliding Window Heap (Priority Queue) Monotonic Queue

Problem - Sliding Window Maximum

Hard

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]

 

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104
  • 1 <= k <= nums.length

Solutions

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class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        # queue = [(-val, i) for i, val in enumerate(nums[: k - 1])]
        # heapify(queue)
        # result = []

        # for i in range(k - 1, len(nums)):
        #     heappush(queue, (-nums[i], i))
        #     while queue[0][1] <= i - k:
        #         heappop(queue)
        #     result.append(-queue[0][0])

        # return result

        queue = deque()
        result = []
        for i, val in enumerate(nums):
            if queue and i - queue[0] >= k:
                queue.popleft()

            while queue and nums[queue[-1]] <= val:
                queue.pop()
            queue.append(i)

            if i >= k - 1:
                result.append(nums[queue[0]])

        return result

Submission Stats:

  • Runtime: 159 ms (71.17%)
  • Memory: 32.1 MB (21.14%)