<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(object):
    def minOperationsToMakeMedianK(self, nums, k):
        # Initialize the total number of operations needed to 0
        ans = 0
        
        # Get the number of elements in the list
        n = len(nums)
        
        # Calculate the index of the median
        mid = n // 2
        
        # Sort the list to facilitate median manipulation
        nums.sort()
        
        # Check if the current median is greater than k
        if nums[mid] > k:
            # If the median is greater than k, we need to reduce the values
            # to make the median equal to k
            while mid >= 0:
                # Stop if we encounter a value less than or equal to k
                if nums[mid] <= k: break
                # Add the absolute difference between the current number and k
                # to the total operations
                ans += abs(k - nums[mid])
                # Move to the previous element
                mid -= 1
        else:
            # If the median is less than or equal to k, we need to increase
            # the values to make the median equal to k
            while mid < n:
                # Stop if we encounter a value greater than or equal to k
                if nums[mid] >= k: break
                # Add the absolute difference between the current number and k
                # to the total operations
                ans += abs(k - nums[mid])
                # Move to the next element
                mid += 1
        
        # Return the total number of operations needed to make the median equal to k
        return ans
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