# 153. Find Minimum in Rotated Sorted Array

## Medium

***

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.

```python
class Solution:
    def findMin(self, nums: List[int]) -> int:
        pivot = self.findPivot(nums)
        return nums[pivot]
        
    def findPivot(self, nums):
        low = 0
        high = len(nums) - 1
        if low == high:
            return low
        elif nums[low] < nums[high]:
            return low
        while low <= high:
            mid = low +(high-low)//2
            if mid < high and nums[mid] > nums[mid+1]:
                return mid+1
            elif mid > low and nums[mid-1]>nums[mid]:
                return mid
            elif nums[mid] > nums[low]:
                low = mid+1
            else:
                high = mid-1
            
```

Solution is same as for PB 154 : Find Minimum in Rotated Sorted Array II

```python
class Solution:
    def findMin(self, nums: List[int]) -> int:
        left , right = 0, len(nums)-1
        while left < right:
            mid = left + (right-left)//2
            if nums[mid] > nums[right]:
                left = mid + 1
            elif nums[mid] < nums[right]:
                right = mid
            else:
                right -= 1
        return nums[left]
            
```