<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:
    def minSwaps(self, nums: List[int]) -> int:
        """
        A way of removing the circular array property from an array
        is by appending it to itself.

        The solution to this problem is simply finding a sub-array of
        size `m` that encloses all the `1`s in the original array and
        the number of `0`s in that sub-array will be the result.

        - [0, 1, 1, 0, 0, 1, 0] -> n = 7, one_count = 3
        - [0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0] -> "de-circularize" the array
        
        Do a sliding window of size `m <= n` that includes `one_count` ones 
        and count the zeroes present in the window.

        - [0, {1, 1, 0, 0, 1}, 0, 0, 1, 1, 0, 0, 1, 0] -> r = 2
        - [0, 1, {1, 0, 0, 1, 0, 0, 1}, 1, 0, 0, 1, 0] -> r = 4
        - [0, 1, 1, 0, 0, {1, 0, 0, 1, 1}, 0, 0, 1, 0] -> r = 2
        - [0, 1, 1, 0, 0, 1, 0, 0, {1, 1, 0, 0, 1}, 0] -> r = 2
        """
        minimum_swaps = float("inf")

        total_ones = sum(nums)

        ones_count = nums[0]
        end = 0
        for start in range(len(nums)):
            # Remove the element that is left out of the window
            if start != 0:
                ones_count -= nums[start - 1]

            # Expand the window until it fits the total number of ones
            while end - start + 1 < total_ones:
                end += 1
                # This module "turns around" in the circular array
                ones_count += nums[end % len(nums)]

            minimum_swaps = min(minimum_swaps, total_ones - ones_count)

        return minimum_swaps
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