<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 {
public:
    int minMoves(vector<int>& nums, int k) {
        vector<int> temp;
        for(int i = 0; i < nums.size(); i++) if(nums[i] == 1) temp.push_back(i);
        for(int i = 0; i < temp.size(); i++) temp[i] -= i;
        
        int median = (k - 1)/2;
        int ans = INT_MAX, l_sum = 0, r_sum = 0;
        int l, r;
        if(k % 2) l = r = (k/2);
        else {
            l = (k/2 - 1);
            r = k/2;
        }
        
        for(int i = 0; i < median; i++) l_sum += (temp[median] - temp[i]);
        for(int i = median+1; i < k; i++) r_sum += (temp[i] - temp[median]);
        ans = (l_sum + r_sum);
        for(int i = k; i < temp.size(); i++){
            l_sum -= temp[median] - temp[i - k];
            l_sum += l*(temp[median+1] - temp[median]);
            r_sum += temp[i] - temp[median + 1];
            r_sum -= r*(temp[median+1]-temp[median]);
            ans = min(ans, l_sum + r_sum);
            median++;
        }
        return ans;
    }
};
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