<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 op1 = minSwapsHelper(nums, 0);  // Grouping all 0s together

        int op2 = minSwapsHelper(nums, 1);  // Grouping all 1s together

        return min(op1, op2);

    }

private:

    int minSwapsHelper(const vector<int>& data, int val) {

        int length = data.size();

        vector<int> rightSuffixSum(length + 1, 0);

        // Construct the suffix sum array for counting opposite values

        // (val ^ 1)

        for (int i = length - 1; i >= 0; i--) {

            rightSuffixSum[i] = rightSuffixSum[i + 1];

            if (data[i] == (val ^ 1)) rightSuffixSum[i]++;

        }

        // Total zeros in the array if `val` is 1 (or vice versa)

        int totalSwapsNeeded = rightSuffixSum[0];

        // Track current number of required swaps in the current segment

        int currentSwapCount = 0;

        int minimumSwaps =

            totalSwapsNeeded - rightSuffixSum[length - totalSwapsNeeded];

        // Iterate to find the minimum swaps by sliding the potential block of

        // grouped `val`

        for (int i = 0; i < totalSwapsNeeded; i++) {

            if (data[i] == (val ^ 1)) currentSwapCount++;

            int remaining = (totalSwapsNeeded - i - 1);

            int requiredSwaps =

                ((i + 1) - currentSwapCount) +

                (remaining - rightSuffixSum[length - remaining]);

            minimumSwaps = min(minimumSwaps, requiredSwaps);

        }

        return minimumSwaps;

    }

};





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