LeetCode 33. Search in Rotated Sorted Array (javascript)

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

1 <= nums.length <= 5000
-10⁴ <= nums[i] <= 10⁴
All values of nums are unique.
nums is guaranteed to be rotated at some pivot.
-10⁴ <= target <= 10⁴

Idea:

Binary Search: the idea is to create a recursive function that takes l and h as range in input and the key.

1) Find middle point mid = (l + h)/2
2) If key is present at middle point, return mid.
   Edge case check(not found): if (l > h) return -1;
3) Else If nums[l..mid] is sorted
    a) If key to be searched lies in range from nums[l]
       to nums[mid], recur for nums[l..mid].
    b) Else recur for nums[mid+1..h]
4) Else (nums[mid+1..h] must be sorted)
    a) If key to be searched lies in range from nums[mid+1]
       to nums[h], recur for nums[mid+1..h].
    b) Else recur for nums[l..mid]

Solution:

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    return searching(nums, 0, nums.length - 1, target);
};

function searching(nums, l, h, target) {
    // basic case: not found
    if (l > h) return -1;
    
    let mid = Math.floor((l + h) / 2);
    // basic case: found
    if (target === nums[mid]) return mid;
    
    // if nums[l...mid] is sorted
    if (nums[l] <= nums[mid]) { /// make sure is <=
        if (target >= nums[l] && target <= nums[mid])
            return searching(nums, l, mid - 1, target);
        else
            return searching(nums, mid + 1, h, target)
    }
    
    // if nums[l...mid] is notesorted, then nums[mid+1..h] must be sorted
    if (target >= nums[mid] && target <= nums[h])
        return searching(nums, mid + 1, h, target);
    else
        return searching(nums, l, mid - 1, target);
}