153. Find Minimum in Rotated Sorted Array

1. Description

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.

2. Example

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.

3. 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.

4. Solutions

Binary Search

n = nums.size()
Time complexity: O(logn)
Space complexity: O(1)

class Solution {
public:
    int findMin(const vector<int> &nums) {
        if (nums.front() <= nums.back()) {
            return nums.front();
        }

        const int n = nums.size();
        int left = 0, right = n;
        while (left < right) {
            int middle = left + (right - left) / 2;
            if (nums[middle] >= nums.front()) {
                left = middle + 1;
            } else {
                right = middle;
            }
        }

        return nums[left];
    }
};