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 O(log n) time. (Problem)

				
    class Solution:
        def findMin(self, nums: List[int]) -> int:
            res=nums[0]
            l,r=0,len(nums)-1
            
            while(l<=r):
                if(nums[l]<=nums[r]): 
                    res= min(res,nums[l])
                
                mid= (l+r)//2
                res=min(nums[mid],res)
                
                if(nums[mid]>=nums[l]):
                    l=mid+1
                else:
                    r=mid-1
            
            return res
							
			

Time Complexity => o(log n)

Space Complexity => o(1)