Minimum Adjacent Swaps for K Consecutive Ones
<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>
Greedy; Array; Prefix Sum; Sliding Window
func minMoves(nums []int, k int) int {
	arr := []int{}
	for i, x := range nums {
		if x != 0 {
			arr = append(arr, i)
		}
	}
	s := make([]int, len(arr)+1)
	for i, x := range arr {
		s[i+1] = s[i] + x
	}
	ans := 1 << 60
	x := (k + 1) / 2
	y := k - x
	for i := x - 1; i < len(arr)-y; i++ {
		j := arr[i]
		ls := s[i+1] - s[i+1-x]
		rs := s[i+1+y] - s[i+1]
		a := (j+j-x+1)*x/2 - ls
		b := rs - (j+1+j+y)*y/2
		ans = min(ans, a+b)
	}
	return ans
}