<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>

using LL = long long;
class Solution {
public:
    int minMoves(vector<int>& nums, int k) {
        vector<int> ones;
        for(int i=0; i<nums.size(); i++) if(nums[i]==1) ones.push_back(i);
        int n = ones.size();
        LL offset = 0;
        for(int i=0; i<k; i++) offset += abs(i-k/2);
        LL min_move = LONG_LONG_MAX;
        LL cur_move = 0;
        int l = 0;
        int r = 0;
        while(r<n){
            cur_move += ones[r] - ones[(l+r)/2];
            if(r>=k){
                cur_move += ones[l] - ones[(l+1+r)/2];
                l++;
            }
            if(r>=k-1) min_move = min(min_move, cur_move);
            r++;
        }
        return min_move - offset;
    }
};
131900