<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) {
        
        if(k==1){
            return 0;
        }
        vector<int>freq;
        int n=nums.size();

        int f=0;
        int idx=0;
        
        for(int i=0;i<n;i++){
            if(nums[i]==1){
                idx=i;
                break;
            }
        }

        for(int i=idx+1;i<n;i++){
            if(nums[i]==1){
                freq.push_back(f);
                f=0;
            }
            else{
                f++;
            }
        }
       
        n=freq.size();
       
        vector<int>pf(n);
        pf[0]=freq[0];
        for(int i=1;i<n;i++){
            pf[i]=freq[i];
            pf[i]+=pf[i-1];
        }
        
        if(k%2==0){

            idx=0;
            int ans=0;
            for(int i=1;i<=k/2;i++){
                ans+=(i*freq[idx]);
                idx++;
            }
            for(int i=k/2-1;i>=1;i--){
                ans+=(i*freq[idx]);
                idx++;
            }
            int ca=ans;
            idx;
            int p1=0;

            int p2=k/2-1;
            for(int i=idx;i<n;i++){
                ca-=freq[p1];
                ca+=freq[i];
                ca-=(pf[p2]-pf[p1]);
                ca+=(pf[i-1]-pf[p2]);
                p2++;
                p1++;
                ans=min(ans,ca);
            }

            return ans;

        }
        else{
            idx=0;
            int ans=0;
            for(int i=1;i<=k/2;i++){
                ans+=(i*freq[idx]);
                idx++;
            }
            for(int i=k/2;i>=1;i--){
                ans+=(i*freq[idx]);
                idx++;
            }
            int ca=ans;
           
            int p1=0;
            int p2=k/2;
            for(int i=idx;i<n;i++){
                ca-=freq[p1];
                ca+=freq[i];
                ca-=(pf[p2]-pf[p1]);
                ca+=(pf[i-1]-pf[p2-1]);
                p2++;
                p1++;
                ans=min(ans,ca);
            }

            return ans;
        }
    }
};
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