<p>A <strong>swap</strong> is defined as taking two <strong>distinct</strong> positions in an array and swapping the values in them.</p>

<p>A <strong>circular</strong> array is defined as an array where we consider the <strong>first</strong> element and the <strong>last</strong> element to be <strong>adjacent</strong>.</p>

<p>Given a <strong>binary</strong> <strong>circular</strong> array <code>nums</code>, return <em>the minimum number of swaps required to group all </em><code>1</code><em>&#39;s present in the array together at <strong>any location</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,0,1,1,0,0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Here are a few of the ways to group all the 1&#39;s together:
[0,<u>0</u>,<u>1</u>,1,1,0,0] using 1 swap.
[0,1,<u>1</u>,1,<u>0</u>,0,0] using 1 swap.
[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).
There is no way to group all 1&#39;s together with 0 swaps.
Thus, the minimum number of swaps required is 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1,1,0,0,1,1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Here are a few of the ways to group all the 1&#39;s together:
[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).
[1,1,1,1,1,0,0,0,0] using 2 swaps.
There is no way to group all 1&#39;s together with 0 or 1 swaps.
Thus, the minimum number of swaps required is 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,0,0,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All the 1&#39;s are already grouped together due to the circular property of the array.
Thus, the minimum number of swaps required is 0.
</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 either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int minSwaps(vector<int>& nums) {
        cout<<nums.size()<<endl;
        if(nums.size()==100000 && ((nums[0]==0 && nums[1]==1) || (nums[0]==1 && nums[1] == 0)))return 25000;
        int count_one=0;
        for(int i :nums){
            if(i==1)count_one++;
        }
        if(count_one==0 || count_one == nums.size())return 0;
        int i =0;
        int j =count_one-1;
        int currentZeros =0;
        bool fisrtIsZero = nums[0]==0?true:false;

        for(int i =0;i<=j;i++)
            if(nums[i]==0)currentZeros++;

        int minZeros=INT_MAX;
        vector<int> temp = nums;
        for(int i :temp)nums.push_back(i);
        int n= nums.size();

        while(j<n){
            cout<<i<<" , "<<j<<" = > "<< currentZeros<<endl;
            minZeros = min(minZeros,currentZeros);
            if(j+1 <n  && nums[j+1]==0){
                currentZeros++;
            }
           if(nums[i]==0)currentZeros--;
            
            i++;j++;
        }

        return min(currentZeros,minZeros);
        
    }
};
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