# 153 Find Minimum in Rotated Sorted Array

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:

``````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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 `````` ``````class Solution: def findMin(self, nums: List[int]) -> int: if len(nums) == 1: return nums def helper(l: int, r: int) -> int: if nums[r] > nums[l]: return nums[l] mid = l + (r - l) // 2 if nums[mid] > nums[mid + 1]: return nums[mid + 1] if nums[mid-1] > nums[mid]: return nums[mid] if nums[mid] > nums[l]: return helper(mid+1, r) else: return helper(l, mid-1) return helper(0, len(nums) - 1)``````