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# LeetCode 153. Find Minimum in Rotated Sorted Array

Binary Search

Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` 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*.

**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

[English Version in Youtube](https://youtu.be/sNP3mg6dGAU)

[中文版解答Youtube Link](https://youtu.be/PTS5CXXBT1s)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV15h411X72i/)

```
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 {
                i = mid + 1;
            }
        }
        
        return nums[i];
    }
};
```