<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;
        }
        int n = nums.size();
        
        // store index of all one in sorted order
        vector<int> index;
        for(int i=0;i<n;i++){
            if(nums[i]){
                index.push_back(i);
            }
        }

        // calculate prefix sum of index array
        vector<long long> prefix;
        long long sum = 0;
        for(int x : index){
            sum += x;
            prefix.push_back(sum);
        }

        // now take window of k size consecutively
        int left = 0, right = k-1;
        long long radius = (k%2 ? k/2 : (k-2)/2);
        long long ans = INT_MAX;
        while(right < index.size()){
            int pivot = (left + right)/2;
            long long temp = 0;
            if(k%2){
                temp = prefix[right]-prefix[pivot] -
                        ((pivot > 0 ? prefix[pivot-1] : 0)-(left >0 ? prefix[left-1] : 0)) -
                        radius*(radius+1);
            }
            else{
                temp = prefix[right]-prefix[pivot] -
                        ((pivot>0 ? prefix[pivot-1] : 0)-(left > 0 ? prefix[left-1] : 0)) - index[pivot] -
                        radius*(radius+1) - (radius+1);
            }
            ans = min(ans, temp);
            left++; right++;
        }

        return ans;
    }
};
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