<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 n = nums.size();
        int zc = 0, oc = 0;
        for (auto it : nums) {
            if (it == 0)
                zc++;
            else
                oc++;
        }
        if (n == 1 || zc == n || oc == n)
            return 0;
        vector<int> numzeros(n, 0);
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0)
                cnt++;
            numzeros[i] = cnt;
        }
        //for (auto it : numzeros) cout << it << " ";
        //cout << endl;
        int mins = zc - numzeros[zc - 1];
        for (int i = zc; i < n; i++) {
            mins = min(mins, zc - (numzeros[i] - numzeros[i - zc]));
            if (i >= (n - zc + 1)) {
                //cout << numzeros[n - 1] - numzeros[i - 1] << " " << numzeros[zc - (n - i) - 1] << endl;
                mins = min(mins, (zc - ((numzeros[n - 1] - numzeros[i - 1]) + numzeros[zc - (n - i) - 1])));
            }
        }
        return mins;
    }
};
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