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LeetCode Solutions

239. Sliding Window Maximum

Time: $O(n)$

Space: $O(n)$

			


class Solution {
 public:
  vector<int> maxSlidingWindow(vector<int>& nums, int k) {
    vector<int> ans;
    deque<int> q;  // Max queue

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

    return ans;
  }
};




			


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

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

    return ans;
  }
}




			


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

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

    return ans