Sliding Window Maximum - Leetcode 239

Sliding Window Maximum - Leetcode 239


4 min read

Problem - 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]
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]


  • 1 <= nums.length <= 10<sup>5</sup>

  • -10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>

  • 1 <= k <= nums.length

Solution in Golang

func maxSlidingWindow(nums []int, k int) []int {
    output := []int{}
    q := make([]int, 0)
    l, r := 0, 0
    for r < len(nums) {
        for len(q) != 0 && nums[q[len(q)-1]] < nums[r] {
            q = q[:len(q)-1]
        q = append(q, r)
        if l > q[0] {
            q = q[1:]
        if (r + 1) >= k {
            output = append(output, nums[q[0]])
    return output

This code defines a function maxSlidingWindow that takes two parameters: a slice of integers nums and an integer k. The goal of this function is to find the maximum element in a sliding window of size k as it moves from left to right through the nums slice and return the maximum values in a new slice.

Here's a step-by-step explanation of how the code works:

  1. Initialize output as an empty slice of integers to store the maximum values found in the sliding window.

  2. Create an empty slice q to act as a deque (double-ended queue) for storing indices of elements in the nums slice.

  3. Initialize two pointers l and r to 0. l represents the left end of the sliding window, and r represents the right end of the sliding window.

  4. Start a for loop that continues until the right pointer r reaches the end of the nums slice.

  5. Inside the loop, there is another for loop that runs while the deque q is not empty and the element at the last index in q (i.e., nums[q[len(q)-1]]) is less than the current element nums[r]. This inner loop removes elements from the back of the deque q until the condition is met, effectively maintaining a deque of decreasing elements.

  6. After the inner loop, the current index r is appended to the deque q. This is done because it is possible that the maximum element for the current window might be the element at index r.

  7. Check if the left pointer l is greater than the index stored at the front of the deque q. If it is, this means that the front element in q is outside the current window, so we remove it from the front of the deque.

  8. Check if the size of the current sliding window (i.e., r+1) is greater than or equal to k. If it is, this means that the window has reached the desired size of k, and we can add the maximum element in the window (which is nums[q[0]], where q[0] stores the index of the maximum element) to the output slice. Then, increment the left pointer l.

  9. Increment the right pointer r to move the sliding window one step to the right.

  10. Repeat steps 4 to 9 until the right pointer r reaches the end of the nums slice.

  11. Finally, return the output slice, which contains the maximum values for each sliding window of size k.

In summary, this code efficiently finds the maximum values in a sliding window of size k as it moves through the input slice nums using a deque data structure to optimize the process.

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