Problem Statement:

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.

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:

Notice that to verify if an array element is minimum or not, it is sufficient to verify that it is smaller than both its neighbors (with rotation).ie

 
prev = (mid==0) ? n-1 : mid-1
next = (mid==n-1) ? 0 : mid+1
A[mid]<=A[prev] and A[mid]<=A[next]

We want to find an index mid in the range [0,n) that satisfies this condition. We use binary search for this.

 
int findMin(vector<int>& nums) 
{
	int n=nums.size(), lo=0, hi=n-1, mid;
	while (lo<=hi)
	{
		mid = lo + (hi-lo)/2;
		int prev = (mid-1+n)%n, next = (mid+1)%n;
		if (nums[mid]<=nums[prev] && nums[mid]<=nums[next])
			break;
		else if (nums[mid] <= nums[hi])
			hi = mid-1;
		else
			lo = mid+1; // nums[0...mid] subarray is sorted
	}
	return nums[mid];
}