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.

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> &nums)
  {
    int n = nums.size(), s = 0, e = n - 1;
    int ans = 0;
    while (s <= e)
    {
      int mid = (s + e) / 2;
      if (mid + 1 < n && mid - 1 >= 0 && nums[mid] < nums[mid + 1] && nums[mid] < nums[mid - 1])
        return nums[mid];
      else if (nums[mid] > nums[e])
        s = mid + 1;
      else
        e = mid - 1;
    }
    return nums[s];
  }
};