<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 n=nums.size();
        queue<int> q1;
        queue<int> q2;
        long long sum1=0;
        long long sum2=0;
        vector<long long> left(n,-1);
        vector<long long> right(n,-1);
        for(int i=0;i<n;i++)
        {
            if(nums[i]==1)
            {
                sum1+=i;
                q1.push(i);
                if(q1.size()>((k+1)/2))
                {
                    sum1-=q1.front();
                    q1.pop();
                }
                if(q1.size()==((k+1)/2))
                {
                    left[i]=sum1;
                }
            }
            if(nums[n-i-1]==1)
            {
                sum2+=(n-i-1);
                q2.push(n-i-1);
                if(q2.size()>((k+1)/2))
                {
                    sum2-=q2.front();
                    q2.pop();
                }
                if(q2.size()==((k+1)/2))
                {
                    right[n-i-1]=sum2;
                }
            }
        }
        long long mini=1e16;
        if((k%2)==0)
        {
            int j=0;
            for(int i=0;i<n;i++)
            {
                if(left[i]!=-1)
                {
                    while(j<=i)
                    {
                        j++;
                    }
                    while(j<n&&right[j]==-1)
                    {
                        j++;
                    }
                    if(j<n)
                    {
                        mini=min(mini,right[j]-left[i]);
                    }
                }
            }
        }
        else
        {
            for(int i=0;i<n;i++)
            {
                if(left[i]!=-1&&right[i]!=-1)
                {
                    mini=min(mini,right[i]-left[i]);
                }
            }
        }
        long long half1=(k/2);
        long long half2=((k-1)/2);
        return mini-((half1*(half1+1))/2)-((half2*(half2+1))/2);
    }
};
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