<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:
        ones = [i for i in range(len(nums)) if nums[i]]

        # get initial cost of first window
        min_swaps = 0
        median = (k-1) // 2
        for i in range(k):
            min_swaps += abs(ones[median] - ones[i]) - abs(median - i)

        cost = min_swaps
        l = 0
        for r in range(k, len(ones)):
            # subtract leftmost element
            cost -= abs(ones[median] - ones[l]) - abs(median - l)
            l+=1
            new_median = (l + r) // 2
            
            if new_median != median:
                diff = ones[new_median] - ones[median] - 1
                cost += (new_median - l) * diff - (r - new_median) * diff
                # print("cost change", cost)

            cost += abs(ones[new_median] - ones[r]) - abs(new_median - r)
            # print("new added cost", cost)
            median = new_median
            min_swaps = min(min_swaps, cost)

        return min(min_swaps, cost)
23400