<p>You are given an integer array <code>nums</code> and a <strong>non-negative</strong> integer <code>k</code>. In one operation, you can increase or decrease any element by 1.</p>

<p>Return the <strong>minimum</strong> number of operations needed to make the <strong>median</strong> of <code>nums</code> <em>equal</em> to <code>k</code>.</p>

<p>The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,5,6,8,5], k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>We can subtract one from <code>nums[1]</code> and <code>nums[4]</code> to obtain <code>[2, 4, 6, 8, 4]</code>. The median of the resulting array is equal to <code>k</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,5,6,8,5], k = 7</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>We can add one to <code>nums[1]</code> twice and add one to <code>nums[2]</code> once to obtain <code>[2, 7, 7, 8, 5]</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,5,6], k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>The median of the array is already equal to <code>k</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minOperationsToMakeMedianK(self, nums: List[int], k: int) -> int:
        n = len(nums)
        max_pq = []
        min_pq = []
        for num in nums:
            if num < k:
                heapq.heappush(max_pq, -num)
            else:
                heapq.heappush(min_pq, num)

        rt_size = (n - 1) // 2 + 1
        ans = 0
        sz = len(min_pq)
        
        while sz > rt_size:
            front = min_pq[0]
            heapq.heappop(min_pq)

            ans += front - k

            # heapq.heappush(max_pq, k)
            sz -= 1

        pushed = False
        while sz < rt_size:
            front = max_pq[0]
            heapq.heappop(max_pq)

            ans += (k - -front)

            if not pushed:
                pushed = True
                heapq.heappush(min_pq, k)
            sz += 1



        # 加min_pq[0] - k因为min_pq最小项不一定是k
        return ans + (min_pq[0] - k)
37300