<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) {
        int i,ka=0,p,q,j,mid,ans=INT_MAX;
        vector<int> s(nums.size()+1,0),v;
        s[0]=nums[0];
        if(nums[0]==1) v.push_back(0);
        for(i=1;i<nums.size();i++){
            s[i]=s[i-1]+nums[i];
            if(nums[i]==1) v.push_back(i);
        }
        p=k/2;
        q=k/2;
        if(k%2==0) p--;
        mid=v[p];
        for(i=0;i<p;i++){
             int x=mid-v[i];
             int y=s[mid]-s[v[i]];
             x-=y;
             ka+=x;
        }
        // cout<<i+1<<endl;
        for(i=p+1;i<=p+q;i++){
             int x=v[i]-mid;
             int y=s[v[i]]-s[mid];
             x-=y;
             ka+=x;
        }
        // cout<<i+1<<endl;
        ans=min(ans,ka);
        j=i;
        i=1;
        cout<<j<<" "<<ans<<endl;
        // cout<<p<<" "<<q<<endl;
        while(j<v.size()){
            ka+=(v[j]-v[p+1]-1);
            ka-=(v[p]-v[i-1]-1);
            if(k%2==0) ka-=(v[p+1]-v[p]);
            ans=min(ans,ka);
            cout<<j<<" "<<ans<<endl;
            p++;
            i++;
            j++;
        }
        return ans;
    }
};
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