## Problem Statement

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, a, a, ..., a[n-1]]` 1 time results in the array `[a[n-1], a, a, a, ..., 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:

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

Example 2:

 ``````1 2 3 `````` ``````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:

 ``````1 2 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.

See the problem

## Solution

Python3 Solution:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 `````` `````` class Solution: def rotation_point(self, nums): n = len(nums) start = 0 end = n - 1 if n == 1: return 0 while True: mid = (start + end) // 2 if ( nums[mid] <= nums[(mid + 1 + n) % n] and nums[mid] < nums[(mid - 1 + n) % n] ): return mid elif nums[mid] > nums[end]: start = mid + 1 else: end = mid - 1 def findMin(self, nums: List[int]) -> int: i = self.rotation_point(nums) return nums[i] ``````

Test Cases

 `````` 1 2 3 4 5 6 7 8 9 10 11 `````` ``````[3,4,5,1,2] [3,4,5,6, 7, 0,1, 2]  [1,2, 3, 4] [5,1,2,3,4] [2,3,4,5,1] [1,2] [2,1] [1,2,3] [3,1,2] [2,3,1] ``````