<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) {
        vector<int> lr {};
    
        int num_1s = 0;
        for (auto x : nums) {
            if (x == 1) num_1s++;
        }

        // cout << num_1s << "\n";
        if (nums.size() == 0 || num_1s == 0 || num_1s == nums.size()) return 0;

        for (int i = 0; i < nums.size(); ++i) {
            lr.push_back(0);
            // cout << "starting index " << i << endl;
            if (lr.size() > 1) {
                lr[i] = lr[i - 1];
                if (nums[(i - 1) % nums.size()] == 0) lr[i]--;
                if (nums[(i + num_1s - 1) % nums.size()] == 0) lr[i]++;
                // cout << "prev caltuation is " << lr[i - 1] << ", updated to " << lr[i] << endl;
            } else {
                for (int j = i; j != (i + num_1s) % nums.size();) {
                    // cout << "sub index " << j << endl;
                    if (nums[j] == 0) {
                        ++lr[i];
                        // cout << "found 0, starting index " << i << " now at " << lr[i] << endl;
                    }
                    j = (j + 1) % nums.size();
                }
            }
        }

        int min = lr[0];
        for (auto t : lr) {
            if (t < min) min = t;
        }

        return min;
    }
};
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