# 154. Find Minimum in Rotated Sorted Array II

#### Hard

***

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

* `[4,5,6,7,0,1,4]` if it was rotated `4` times.
* `[0,1,4,4,5,6,7]` 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` that may contain **duplicates**, return *the minimum element of this array*.

You must decrease the overall operation steps as much as possible.

 

**Example 1:**

```
Input: nums = [1,3,5]
Output: 1
```

**Example 2:**

```
Input: nums = [2,2,2,0,1]
Output: 0
```

 

**Constraints:**

* `n == nums.length`
* `1 <= n <= 5000`
* `-5000 <= nums[i] <= 5000`
* `nums` is sorted and rotated between `1` and `n` times.

&#x20;

**Follow up:** This problem is similar to [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/), but `nums` may contain **duplicates**. Would this affect the runtime complexity? How and why?

&#x20;Same Solution as it was for Problem 153:

```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]
```