<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:
    long long temp;
    long long ans=LLONG_MAX,mid,len1,len2,val1,val2,x;
    vector<long long> A;
    int minMoves(vector<int>& nums, int k) {
        for(int i=0;i<=nums.size()-1;i++)
            if(nums[i]==1) A.push_back(i);
        vector<long long> prefix(A.size());
        prefix[0]=A[0];
        for(int i=1;i<=A.size()-1;i++)
        {
            prefix[i]=prefix[i-1]+A[i];
        }
        for(int i=k-1;i<=A.size()-1;i++)
        {
            mid=(i+i-k+1)/2;
            len1=mid-(i-k+1)+1;
            len2=k-len1;
            x=A[mid]-len1+1;
            temp=(i-k+1==0)? prefix[mid] : prefix[mid]-prefix[i-k];
            val1=x-(temp-len1*(x+x+len1-1)/2);
            val2=(prefix[i]-prefix[mid]-len2*(x+len1+x+len2+len1-1)/2)-x;
            if(ans>val1+val2) ans=val1+val2;
        }
        return ans;
    }
};
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