## Problem

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.

## Solution

Python3 Solution:

```
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

```
[3,4,5,1,2]
[3,4,5,6, 7, 0,1, 2]
[1]
[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]
```