<p>The <strong>median</strong> is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.</p>

<ul>
	<li>For examples, if <code>arr = [2,<u>3</u>,4]</code>, the median is <code>3</code>.</li>
	<li>For examples, if <code>arr = [1,<u>2,3</u>,4]</code>, the median is <code>(2 + 3) / 2 = 2.5</code>.</li>
</ul>

<p>You are given an integer array <code>nums</code> and an integer <code>k</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 median array for each window in the original array</em>. Answers within <code>10<sup>-5</sup></code> of the actual value will be accepted.</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> [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
<strong>Explanation:</strong> 
Window position                Median
---------------                -----
[<strong>1  3  -1</strong>] -3  5  3  6  7        1
 1 [<strong>3  -1  -3</strong>] 5  3  6  7       -1
 1  3 [<strong>-1  -3  5</strong>] 3  6  7       -1
 1  3  -1 [<strong>-3  5  3</strong>] 6  7        3
 1  3  -1  -3 [<strong>5  3  6</strong>] 7        5
 1  3  -1  -3  5 [<strong>3  6  7</strong>]       6
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,2,3,1,4,2], k = 3
<strong>Output:</strong> [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution:
    def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:
        maxHeap = []
        minHeap = []
        
        # Step 1: Heap Initialization
        # When there are 4 elements: we have 2 in maxHeap and 2 in minHeap
        # When there are 5 elements: we have 2 in maxHeap and 3 in minHeap
        for i in range(k):
            heapq.heappush(maxHeap, (-nums[i], i))
        
        for _ in range(k - k//2):
            num, i = heapq.heappop(maxHeap)
            heapq.heappush(minHeap, (-num, i))
        
        # Step 2: Sliding Window and Maintain Heap Balance
        ans = [(-maxHeap[0][0]+minHeap[0][0])/2.0] if k%2 == 0 else [minHeap[0][0]/1.0]
        for i in range(k, len(nums)):
            
            # always add the element to minHeap
            num, j = heapq.heappushpop(maxHeap, (-nums[i], i))
            heapq.heappush(minHeap, (-num, j))
            
            # Now check the inbalance
            if (nums[i-k], i-k) < minHeap[0]:
                # IMPORTANT: `if nums[i-k] < minHeap[0][0]` won't work!
                # EXPLANATION:
                # there could be a number valued at nums[i-k] with a smaller than (i-k) index on top of minHeap
                # after the above maxHeap heappushpop + minHeap heappush
                # when that's the case, it means that (nums[i-k],i-k) is ACTUALLY in minHeap!
                
                # number to be removed in maxHeap, inbalance
                # transfer one element to maxHeap
                num, j = heapq.heappop(minHeap)
                heapq.heappush(maxHeap, (-num, j))
            
            
            while minHeap and minHeap[0][1] <= i-k:
                heapq.heappop(minHeap)
            while maxHeap and maxHeap[0][1] <= i-k:
                heapq.heappop(maxHeap)
            if k%2 == 0:
                ans.append((-maxHeap[0][0]+minHeap[0][0])/2.0)
            else:
                ans.append((minHeap[0][0])/1.0)            
        return ans
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