Find Minimum in Rotated Sorted Array

Problem statement

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: 1Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]Output: 0Explanation: 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: 11Explanation: 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.

My solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    const len = nums.length;
    
    if (len === 1) {
        return nums[0]
    }
    
    let left = 0;
    let right = len - 1;
    
    while (left < right) {
        const mid = Math.floor((right + left) / 2);
        
        const num = nums[mid];
        const lNum = nums[left];
        const rNum = nums[right];
        
        if (num > lNum && num > rNum) {
            if (lNum < rNum) {
                right = mid - 1
            } else {
                left = mid + 1;
            }
        } else {
            if (num > nums[mid + 1]) {
                left = mid + 1
            } else {
                right = mid;
            }
        }
        
    }
    
    return nums[left];
};