<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) {
        int len=nums.size();
        int arr[len];
        int cnt=0;
        memset(arr,0,sizeof(arr));
        if(nums[0]==1){
                arr[0]=1;
                cnt++;
        }
        for(int i=1;i<len;i++){
            arr[i]=arr[i-1];
            if(nums[i]==1){
                arr[i]++;
                cnt++;
            }
        }

        //cout<<cnt<<endl;

        int mx=len-cnt;
        if(cnt==0)return 0;
        for(int i=0;i<len;i++){
            int ran=i+cnt-1;
            int chk;
            if(i==0)chk=cnt-arr[ran];
            else if(ran>=len){
                    ran=ran-len;
                    int k=arr[len-1]-arr[i-1];
                    chk=cnt-(arr[ran]+k);
            }
            else
                chk=cnt-(arr[ran]-arr[i-1]);

            //cout<<arr[ran]<<" "<<arr[i-1]<<endl;
            mx=min(mx,chk);
            //cout<<i<<" "<<chk<<endl;
        }
        return mx;
    }
};
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