<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 totalOnes = 0;
        vector<int> prefixZeroes(n, 0);
        int csum = 0;

        for(int i = 0; i < n; i++) {
            if(nums[i] == 1) {
                totalOnes++;
            } else {
                csum++;
            }
            prefixZeroes[i] = csum;
        }

        if(totalOnes == 0 || totalOnes == n) return 0;

        for(int i = 0; i < totalOnes; i++) nums.push_back(nums[i]);

        int ans = INT_MAX;
        int i = totalOnes - 1;

        while(i < n + totalOnes) {
            int exchanges = 0;
            if(i < n) {
                exchanges = prefixZeroes[i] - prefixZeroes[i - totalOnes + 1] + (nums[i - totalOnes + 1] == 0);
            } else {
                exchanges = prefixZeroes[i % n] + prefixZeroes[n - 1];
                if(i - totalOnes + 1 < n) {
                    exchanges += int(nums[i - totalOnes + 1] == 0) - prefixZeroes[i - totalOnes + 1];
                }
            }
            ans = min(ans, exchanges);
            i++;
        }
        
        return ans;
    }
};
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