<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:
    import heapq
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        res = []
        word_freq_in_window = defaultdict(int)
        window_heap = []
        
        #initialize the word_freq_in_window
        for i in range(k):
            word_freq_in_window[nums[i]] += 1
            window_heap.append(-1*nums[i])  # heapq is min heap we want max heap
        
        heapq.heapify(window_heap)
        res.append(-1*window_heap[0])

        window_start = 0
        window_end = k-1
        

        while window_end<len(nums)-1:
            # Removing the start value from freq dict
            word_freq_in_window[nums[window_start]] -= 1
            if word_freq_in_window[nums[window_start]] == 0:
                del word_freq_in_window[nums[window_start]]
            
            while window_heap and -1*window_heap[0] not in word_freq_in_window:
                heapq.heappop(window_heap)
            
            window_start += 1
            window_end += 1
            # insert the new word at the end of the window
            word_freq_in_window[nums[window_end]] += 1
            heapq.heappush(window_heap,-1*nums[window_end])

            res.append(-1*window_heap[0])

        return res    

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