> For the complete documentation index, see [llms.txt](https://zhenchaogan.gitbook.io/leetcode-solution/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-154-find-minimum-in-rotated-sorted-array-ii.md).

# LeetCode 154. Find Minimum in Rotated Sorted Array II

Binary Search

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*.

**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 is the same as [Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/) but with duplicates. Would allow duplicates affect the run-time complexity? How and why?

## Solution

[English Version in Youtube](https://youtu.be/rR5M-Z0R3QE)

[中文版解答Youtube Link](https://youtu.be/w_OIDPaf_eI)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV12p4y1b74Q/)

```
class Solution {
public:
    int findMin(vector<int>& nums) {
        int i = 0;
        int j = nums.size() - 1;
        
        while (i < j) {
            int mid = (i + j) / 2;
            if (nums[mid] < nums[j]) {
                j = mid;
            } else if (nums[mid] > nums[j]) {
                i = mid + 1;
            } else {
                j--;
            }
        }
        
        return nums[i];
    }
};
```