find_min_II.py

# 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


from typing import List
class Solution:
    def findMin(self, nums: List[int]) -> int:
        l = 0
        r = len(nums) - 1

        while l < r:
            m = (l+r)//2
            if nums[m] == nums[r]:
                r -= 1
            elif nums[m] < nums[r]:
                r = m
            else:
                l += 1
        return nums[l]