Leetcode 153. 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.

Solution:

class Solution {
public:
    int findMin(vector<int>& nums) {
        int start = 0;
        int end = nums.size() - 1;
        
        while ( end >= start) {        
            if (start + 1 == end) {
                return nums[start] < nums[end] ? nums[start] : nums[end];
            }
              
            int mid = (start + end) / 2;
            if (nums[start] <= nums[mid] && nums[mid] <= nums[end]) {
                // all are ordered
                return nums[start];
            } else if (nums[start] <= nums[mid]) {
                // left is ordered
                start = mid;    
            } else {
                // right is ordered
                end = mid;
            }
        }
        return 0;
    }
};

Runtime: 9 ms, faster than 24.35% of C++ online submissions for Find Minimum in Rotated Sorted Array.

Memory Usage: 10.1 MB, less than 71.74% of C++ online submissions for Find Minimum in Rotated Sorted Array.