find_min.py

# Suppose an array of length n sorted in ascending order is rotated between 1 and n times.
# For example, the array nums = [0,1,2,4,5,6,7] might become:

# [4,5,6,7,0,1,2] if it was rotated 4 times.
# [0,1,2,4,5,6,7] 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 of unique elements, return the minimum element of this array.

# You must write an algorithm that runs in O(log n) time.

 

# Example 1:

# Input: nums = [3,4,5,1,2]
# Output: 1
# Explanation: The original array was [1,2,3,4,5] rotated 3 times.
# Example 2:

# Input: nums = [4,5,6,7,0,1,2]
# Output: 0
# Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
# Example 3:

# Input: nums = [11,13,15,17]
# Output: 11
# Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
 

# Constraints:

# n == nums.length
# 1 <= n <= 5000
# -5000 <= nums[i] <= 5000
# All the integers of nums are unique.
# nums is sorted and rotated between 1 and n times.


class Solution:
    def findMin(self, nums: List[int]) -> int: # type:ignore
        left = 0
        right = len(nums) - 1

        while left < right:
            mid = (left + right) >> 1
            if nums[mid] < nums[right]:
                right = mid
            else:
                left = mid + 1
        return nums[left]