153.find-minimum-in-rotated-sorted-array.cpp

//  Tag: Array, Binary Search
//  Time: O(logN)
//  Space: O(1)
//  Ref: -
//  Note: Rotated;
//  Video: https://youtu.be/GsecRZC5to4

//  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>& nums) {
        int left = 0;
        int right = nums.size() - 1;

        while (left < right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] > nums[right]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return nums[left];
    }
};