<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>

import heapq
from collections import defaultdict
class Solution:
    def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:

        medians = []
        hash_table = defaultdict(int)
        lo = []  # max heap (inverted min heap)
        hi = []  # min heap

        i = 0  # index of current incoming element being processed

        # Initialize the heaps
        while i < k:
            heapq.heappush(lo, -nums[i])  # use negative values to simulate max heap
            i += 1
        for _ in range(k // 2):
            heapq.heappush(hi, -heapq.heappop(lo))

        while True:
            # Get median of current window
            if k % 2 == 1:
                medians.append(float(-lo[0]))  # max heap root for odd k
            else:
                medians.append((-lo[0] + hi[0]) / 2.0)  # average of roots for even k

            if i >= len(nums):
                break  # break if all elements processed

            out_num = nums[i - k]  # outgoing element
            in_num = nums[i]       # incoming element
            i += 1
            balance = 0            # balance factor

            # Number `out_num` exits window
            balance += -1 if out_num <= -lo[0] else 1
            hash_table[out_num] += 1

            # Number `in_num` enters window
            if lo and in_num <= -lo[0]:
                balance += 1
                heapq.heappush(lo, -in_num)
            else:
                balance -= 1
                heapq.heappush(hi, in_num)

            # Re-balance heaps
            if balance < 0:  # `lo` needs more valid elements
                heapq.heappush(lo, -heapq.heappop(hi))
                balance += 1
            if balance > 0:  # `hi` needs more valid elements
                heapq.heappush(hi, -heapq.heappop(lo))
                balance -= 1

            # Remove invalid numbers that should be discarded from heap tops
            while lo and hash_table[-lo[0]]:
                hash_table[-lo[0]] -= 1
                heapq.heappop(lo)
            while hi and hash_table[hi[0]]:
                hash_table[hi[0]] -= 1
                heapq.heappop(hi)

        return medians

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