Sliding Window Maximum LeetCode Solution | Easy Approach

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Sliding Window Maximum 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

Sliding Window Maximum Solutions

Time: O(n)
Space: O(n)

C++

 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;
  }
};

Java

 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;
  }
}

Python


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

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