# LC 153. Find Minimum in Rotated Sorted Array

### [Problem link](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)

###### tags: `leedcode` `python` `c++` `medium` `Binary Search`

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

- <code>[4,5,6,7,0,1,2]</code> if it was rotated <code>4</code> times.
- <code>[0,1,2,4,5,6,7]</code> if it was rotated <code>7</code> times.

Notice that **rotating** an array <code>[a[0], a[1], a[2], ..., a[n-1]]</code> 1 time results in the array <code>[a[n-1], a[0], a[1], a[2], ..., a[n-2]]</code>.

Given the sorted rotated array <code>nums</code> of **unique** elements, return the minimum element of this array.

You must write an algorithm that runs in<code>O(log n) time.</code>

**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:**

- <code>n == nums.length</code>
- <code>1 <= n <= 5000</code>
- <code>-5000 <= nums[i] <= 5000</code>
- All the integers of <code>nums</code> are **unique** .
- <code>nums</code> is sorted and rotated between <code>1</code> and <code>n</code> times.

## Solution 1 - Binary Search
#### Python

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1

        while left < right:
            mid = (left + right) // 2

            if nums[mid] < nums[right]:
                right = mid
            else:
                left = mid + 1

        return nums[left]

#### C++

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

>### Complexity
>n = nums.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(logn) | O(1) |

## Note
sol1:
每次都跟最後一個相比, 來判斷left, right的區間.
[Ref](https://hackmd.io/@Alone0506/rJd4JUe8C)