Leetcode•Mar 20, 2025
239. Sliding Window Maximum
Hazrat Ali
Leetcode
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]
* Sliding Window Maximum
*
* Given an array 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:
*
* Input: nums = [1,3,-1,-3,5,3,6,7], and 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
* Note:
* You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.
*
* Follow up:
* Could you solve it in linear time?
*/
Solution :
/**
* @param {number[]} nums
* @param {number} k
* @return {number[]}
*/
const maxSlidingWindow = (nums, k) => {
const window = [];
const result = [];
for (let i = 0; i < nums.length; i++) {
if (window.length > 0 && window[0] === i - k) {
window.shift();
}
while (window.length > 0 && nums[window[window.length - 1]] < nums[i]) {
window.pop();
}
window.push(i);
if (i >= k - 1) {
result.push(nums[window[0]]);
}
}
return result;
};