Minimum Swaps to Group All 1's Together II

lcpy gen -n 2134   # by problem number
lcpy gen -s minimum_swaps_to_group_all_1s_together_ii   # by problem name

Input: nums = [0,1,0,1,1,0,0]
Output: 1
Explanation: Here are a few of the ways to group all the 1's together:
[0,0,1,1,1,0,0] using 1 swap.
[0,1,1,1,0,0,0] 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's together with 0 swaps.
Thus, the minimum number of swaps required is 1.

Input: nums = [0,1,1,1,0,0,1,1,0]
Output: 2
Explanation: Here are a few of the ways to group all the 1'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's together with 0 or 1 swaps.
Thus, the minimum number of swaps required is 2.

Input: nums = [1,1,0,0,1]
Output: 0
Explanation: All the 1's are already grouped together due to the circular property of the array.
Thus, the minimum number of swaps required is 0.

class Solution:
    # Time: O(n)
    # Space: O(1)
    def min_swaps(self, nums: list[int]) -> int:
        total = sum(nums)
        n = len(nums)
        if total <= 1 or total == n:
            return 0
        window = sum(nums[:total])
        best = window
        for i in range(1, n):
            window += nums[(i + total - 1) % n] - nums[i - 1]
            best = max(best, window)
        return total - best