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239. Sliding Window Maximum 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> maxSlidingWindow(vector<int>& nums, int k) {
    vector<int> ans;
    deque<int> dq;  // max queue

    for (int i = 0; i < nums.size(); ++i) {
      while (!dq.empty() && dq.back() < nums[i])
        dq.pop_back();
      dq.push_back(nums[i]);
      if (i >= k && nums[i - k] == dq.front())  // out of bound
        dq.pop_front();
      if (i >= k - 1)
        ans.push_back(dq.front());
    }

    return ans;
  }
};

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class Solution {
  public int[] maxSlidingWindow(int[] nums, int k) {
    int[] ans = new int[nums.length - k + 1];
    Deque<Integer> dq = new ArrayDeque<>(); // max queue

    for (int i = 0; i < nums.length; ++i) {
      while (!dq.isEmpty() && dq.peekLast() < nums[i])
        dq.pollLast();
      dq.offerLast(nums[i]);
      if (i >= k && nums[i - k] == dq.peekFirst()) // out of bound
        dq.pollFirst();
      if (i >= k - 1)
        ans[i - k + 1] = dq.peekFirst();
    }

    return ans;
  }
}

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class Solution:
  def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
    ans = []
    dq = collections.deque()  # Max queue

    for i, num in enumerate(nums):
      while dq and dq[-1] < num:
        dq.pop()
      dq.append(num)
      if i >= k and nums[i - k] == dq[0]:  # Out of bound
        dq.popleft()
      if i >= k - 1:
        ans.append(dq[0])

    return ans