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&nbsp;O(log n) time.

&nbsp;
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. 


&nbsp;
Constraints:


	n == nums.length
	1 &lt;= n &lt;= 5000
	-5000 &lt;= nums[i] &lt;= 5000
	All the integers of nums are unique.
	nums is sorted and rotated between 1 and n times.


class Solution:
    def findMin(self, nums: List[int]) -> int:
        if len(nums) == 1 or nums[0] < nums[-1]:
            return nums[0]
        
        l, r = 0, len(nums) - 1
        
        while l <= r:
            mid = l + (r - l) // 2
            
            if mid > 0 and nums[mid - 1] > nums[mid]:
                return nums[mid]
            
			# We compare the middle number and the right index number
			# but why we cannot compare it with the left index number?
            if nums[mid] > nums[r]:
                l = mid + 1
            else:
                r = mid - 1
class Solution {
    public int findMin(int[] nums)
    {
        int min=nums[0];

        for(int i=0;i<nums.length;i++)
        {
            if(min>nums[i])
            {
                min=nums[i];
            }
        }
        return min;
    }
}
class Solution {
public:
    int findMin(vector<int>& nums) {
        int result = -1;
        int first = nums[0];

        // check the array for rotated point when it is obtained break the loop and assign result as rotation point
        for(int i = 1; i<nums.size();i++){
            if(nums[i-1]>nums[i]){
                result = nums[i];
                break;
            }
        }
        // if the array is not sorted return first element
        if(result == -1) {
            return first;
        }
        return result;
    }
};
var findMin = function(nums) {
    let min = Math.min(...nums)
    return min;
};
Find Minimum in Rotated Sorted Array