<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]:
        '''
        it looks like all we need to do is move the window, sorting the range and picking the median.
        if k%2==0, media will be the average of the middle two elements (offset+(k//2), offset+(k//2)-1).
        
        the sorting can be costly, but because we move only 1 step each loop, we can just hold a separate k array
        and keep it sorted all the times, by removing the out-of-window item, and putting the in-window item with bisect.

        this will be O(n*log(k)), instead of O(n*k*log(k)), and extra O(k) space.
        '''
        arr=[x for x in nums[:k]]
        arr.sort()
        L=len(nums)
        result=[]

        def getMedian() -> float:
            if k%2==0:
                return (arr[(-1 + k//2)]+arr[(k//2)])/2
            else:
                return arr[(k-1)//2]
        
        result.append(getMedian())
        i=1
        while i<L-k+1:
            # remove prev item
            p=nums[i-1]
            n=nums[i+k-1]
            arr[bisect_left(arr,p)]=n
            arr.sort()
            # arr.pop(bisect_left(arr,p))
            # arr.insert(bisect_left(arr,n),n)
            result.append(getMedian())
            i+=1
        
        return result
        
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