Count number of times sorted array has been rotated

rotated_sorted_array.py

def num_rotations(nums: list) -> int:
    """
    Returns the number of times a sorted array has been rotated to the right
    with the numbers wrapping to the beginning when they are rotated at the end
    of the list

    >>> nums  = [3, 4, 5, 1, 2]
    >>> num_rotations(nums)
    3
    >>> nums2  = [1, 2, 3, 4, 5]
    >>> num_rotations(nums2)
    0
    >>> nums2  = [4, 5, 1, 2, 3]
    >>> num_rotations(nums2)
    2
    >>> nums3  = [34, 200, 1010, 50982, -1]
    >>> num_rotations(nums3)
    4
    >>> nums4  = [20, 30, -100, -2]
    >>> num_rotations(nums4)
    2
    >>> nums5  = [18]
    >>> num_rotations(nums5)
    0
    """
    min_idx = 0

    # Binary search for minimum: O(log n) time complexity
    def min_finder(nums: list, lower_i: int, upper_i: int) -> int:

        nonlocal min_idx

        if lower_i > upper_i:
            return

        mid = lower_i + (upper_i - lower_i) // 2
        if nums[lower_i] <= nums[mid]:
            min_idx = lower_i if nums[lower_i] < nums[min_idx] else min_idx
            min_finder(nums, mid + 1, upper_i)
        else:
            min_idx = mid if nums[mid] < nums[min_idx] else min_idx
            min_finder(nums, lower_i, mid - 1)

    min_finder(nums, 0, len(nums) - 1)
    return min_idx

Ah, I actually think min_finder (in its current form) is $O(n)$ according to the Master Method because $2$ recursive calls are made whereas binary search only ever makes $1$ recursive call!

EDIT: Fixed recursive implementation to be properly $O(\log n)$.