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239. Sliding Window Maximum

Difficulty Topics

Description

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

Solution

sliding-window-maximum.py
class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        res = []
        queue = collections.deque()

        for i,x in enumerate(nums):
            while queue and x >= nums[queue[-1]]:
                queue.pop()

            queue.append(i)

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

            if queue and i + 1 - k >= queue[0]:
                queue.popleft()

        return res
sliding-window-maximum.cpp
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> deq;
        vector<int> res;

        for (int i = 0; i < nums.size(); i++){
            if (i >= k && deq.front() == nums[i-k])
                deq.pop_front();

            while (deq.size() > 0 && deq.back() < nums[i])
                deq.pop_back();

            deq.push_back(nums[i]);

            if (i >= k-1)
                res.push_back(deq.front());

        }

        return res;
    }
};