LeetCode in TypeScript
239. Sliding Window Maximum
Hard
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]
Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]
Example 4:
Input: nums = [9,11], k = 2
Output: [11]
Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 1041 <= k <= nums.length
Solution
function maxSlidingWindow(nums: number[], k: number): number[] {
const result: number[] = []
// Maintain indices in the deque
const deque: number[] = []
for (let i = 0; i < nums.length; i++) {
// Remove indices that are out of the window
while (deque.length > 0 && deque[0] <= i - k) {
deque.shift()
}
// Remove indices of smaller elements as they are no longer candidates for maximum
while (deque.length > 0 && nums[i] >= nums[deque[deque.length - 1]]) {
deque.pop()
}
// Add the current index to the deque
deque.push(i)
// Add the maximum element for the current window to the result
if (i >= k - 1) {
result.push(nums[deque[0]])
}
}
return result
}
export { maxSlidingWindow }