<p>You are given an integer array, <code>nums</code>, and an integer <code>k</code>. <code>nums</code> comprises of only <code>0</code>&#39;s and <code>1</code>&#39;s. In one move, you can choose two <strong>adjacent</strong> indices and swap their values.</p>

<p>Return <em>the <strong>minimum</strong> number of moves required so that </em><code>nums</code><em> has </em><code>k</code><em> <strong>consecutive</strong> </em><code>1</code><em>&#39;s</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,0,0,1,0,1], k = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong> In 1 move, nums could be [1,0,0,0,<u>1</u>,<u>1</u>] and have 2 consecutive 1&#39;s.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,0,0,0,0,0,1,1], k = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong> In 5 moves, the leftmost 1 can be shifted right until nums = [0,0,0,0,0,<u>1</u>,<u>1</u>,<u>1</u>].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,0,1], k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> nums already has 2 consecutive 1&#39;s.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is <code>0</code> or <code>1</code>.</li>
	<li><code>1 &lt;= k &lt;= sum(nums)</code></li>
</ul>

class Solution:
    def minMoves(self, nums: List[int], k: int) -> int:
        # Step 1: Identify the positions of all 1s in the array
        loc = [i for i, x in enumerate(nums) if x == 1]
        
        # Step 2: Create a prefix sum array for the positions of 1s
        prefix = [0]
        for x in loc:
            prefix.append(prefix[-1] + x)
        
        # Step 3: Initialize the minimum number of moves to infinity
        ans = float('inf')
        
        # Step 4: Iterate over possible starting points for k consecutive 1s
        for i in range(len(loc) - k + 1):
            # Calculate the cost to make k consecutive 1s starting from index i
            # The cost is calculated using the prefix sum array
            ans = min(ans, (prefix[i + k] - prefix[i + (k + 1) // 2]) - (prefix[i + k // 2] - prefix[i]))
        
        # Step 5: Adjust the result to account for the median position
        return ans - (k // 2) * ((k + 1) // 2)
27600