<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 MedianFinder:

    def __init__(self):
        self.low = [] # max heap
        self.high = [] # min heap
        
    def addNum(self, num: int) -> None:
        # low is a max heap, can hold up to n + 1 elements
        # high is a min heap, can hold up to n elements
        heapq.heappush(self.low, -num)
        heapq.heappush(self.high, -heapq.heappop(self.low))
        if len(self.low) < len(self.high):
            heapq.heappush(self.low, -heapq.heappop(self.high))

    def findMedian(self) -> float:
        if len(self.low) == len(self.high):
            return (-self.low[0] + self.high[0]) / 2.0
        return -self.low[0]




class Solution:
    def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:

        low, high = [], nums[:k]
        heapq.heapify(high)
        while len(low) < len(high):
            heapq.heappush(low, -heapq.heappop(high))
        
        removal_cnter = collections.Counter()
        medians = []
        i = k - 1
        while i < len(nums):
            print(i)
            med_tmp = (-low[0] + high[0]) / 2.0 if k % 2 == 0 else (-low[0])
            medians.append(med_tmp)

            i += 1
            if i == len(nums):
                break

            num_2_out = nums[i - k]
            num_2_in = nums[i]

            balance = 0
            if num_2_out <= -low[0]:
                balance -= 1
            else:
                balance += 1
            removal_cnter[num_2_out] += 1

            if low and num_2_in <= -low[0]:
                balance += 1
                heapq.heappush(low, -num_2_in)
            else:
                balance -= 1
                heapq.heappush(high, num_2_in)
            
            if balance < 0:
                balance += 1
                heapq.heappush(low, -heapq.heappop(high))
            if balance > 0:
                balance -= 1
                heapq.heappush(high, -heapq.heappop(low))
            
            while low and removal_cnter[-low[0]]:
                removal_cnter[-low[0]] -= 1
                heapq.heappop(low)

            while high and removal_cnter[high[0]]:
                removal_cnter[high[0]] -= 1
                heapq.heappop(high)
        return medians
                
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