<p>You are given an array of integers&nbsp;<code>nums</code>, there is a sliding window of size <code>k</code> which is moving from the very left of the array to the very right. You can only see the <code>k</code> numbers in the window. Each time the sliding window moves right by one position.</p>

<p>Return <em>the max sliding window</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,-1,-3,5,3,6,7], k = 3
<strong>Output:</strong> [3,3,5,5,6,7]
<strong>Explanation:</strong> 
Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       <strong>3</strong>
 1 [3  -1  -3] 5  3  6  7       <strong>3</strong>
 1  3 [-1  -3  5] 3  6  7      <strong> 5</strong>
 1  3  -1 [-3  5  3] 6  7       <strong>5</strong>
 1  3  -1  -3 [5  3  6] 7       <strong>6</strong>
 1  3  -1  -3  5 [3  6  7]      <strong>7</strong>
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        if not nums or k == 0:
            return []
        
        n = len(nums)
        # Return the original list if window size k is 1
        if k == 1:
            return nums

        # List to store the maximum values of each window
        result = []
        # List to keep indices of elements in the current window
        indices = []
        # Set to quickly check if an index is in the window
        window = set()

        for i in range(n):
            # Remove indices that are out of the current window
            if indices and indices[0] < i - k + 1:
                removed_index = indices.pop(0)
                window.remove(removed_index)

            # Maintain indices in decreasing order based on their values
            while indices and nums[indices[-1]] < nums[i]:
                window.remove(indices.pop())
            
            # Add the current index to both list and set
            indices.append(i)
            window.add(i)
            
            # Append the maximum value of the current window to result
            if i >= k - 1:
                result.append(nums[indices[0]])

        return result
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