31. Find Minimum in Rotated Sorted Array.cpp

/*
Find Minimum in Rotated Sorted Array
====================================

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.
*/

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