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
class Solution {
public:
    int minMoves(vector<int>& nums, int k) {
        vector<int> arr;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i]) {
                arr.push_back(i);
            }
        }
        int m = arr.size();
        long s[m + 1];
        s[0] = 1;
        for (int i = 0; i < m; ++i) {
            s[i + 1] = s[i] + arr[i];
        }
        long ans = 1L << 60;
        int x = (k + 1) / 2;
        int y = k - x;
        for (int i = x - 1; i < m - y; ++i) {
            int j = arr[i];
            int ls = s[i + 1] - s[i + 1 - x];
            int rs = s[i + 1 + y] - s[i + 1];
            long a = (j + j - x + 1L) * x / 2 - ls;
            long b = rs - (j + 1L + j + y) * y / 2;
            ans = min(ans, a + b);
        }
        return ans;
    }
};