LeetCode 153. Find Minimum in Rotated Sorted Array (javascript)

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.

Idea:

Use Binary Search Template

Time Complexity: O(logN)
Space Complexity: O(1)

Solution:

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    // if only have one number
    if (nums.length === 1) return nums[0];
    
    let l = 0;
    let r = nums.length;
    // if this array is sorted
    if (nums[0] < nums[r - 1]) return nums[0];
    
    while (l < r) {
        let mid = l + Math.floor((r - l) / 2);
        
        // if previous element is greater than next element,
        // next element is the smallest
        if (nums[mid] > nums[mid + 1]) 
            return nums[mid + 1];
        if (nums[mid - 1] > nums[mid]) 
            return nums[mid];
        
        // first half is sorted: min is on right side
        if (nums[mid] > nums[0]) 
            l = mid;
        else
            r = mid + 1;
    } 
    return -1;
};

Solution 2:

var findMin = function(nums) {
    return searchMin(nums, 0, nums.length - 1);
};

function searchMin(nums, l, r) {
    // exit recursion:
    // only one or two elements in the array
    if (l + 1 >= r)
        return Math.min(nums[l], nums[r]);
    // this array is sorted
    if (nums[l] < nums[r])
        return nums[l];
    
    let mid = l + Math.floor((r - l) / 2);
    return Math.min(searchMin(nums, l , mid), searchMin(nums, mid + 1, r));
}