"""
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

[4,5,6,7,0,1,4] if it was rotated 4 times.
[0,1,4,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

Example 1:
Input: nums = [1,3,5]
Output: 1

Example 2:
Input: nums = [2,2,2,0,1]
Output: 0

Constraints:
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
nums is sorted and rotated between 1 and n times.

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

analysis:
TC: O(N)
"""
from typing import List


class FindMinimumInRotatedSortedArrayII:
    def findMin(self, nums: List[int]) -> int:
        def helper(s, e):
            if s + 1 >= e:
                return min(nums[s], nums[e])
            if nums[s] < nums[e]:
                return nums[s]
            mid = s + (e - s) // 2
            return min(helper(s, mid - 1), helper(mid, e))

        return helper(0, len(nums) - 1)