Search in Rotated Sorted Array

Problem Description

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

Prior to being passed to your function, nums is possibly left rotated at an unknown index k (1 <= 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 left rotated by 3 indices and become [4,5,6,7,0,1,2].

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

You must write an algorithm with O(log n) runtime complexity.

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 an ascending array that is possibly rotated.
• -104 <= target <= 104

Solution (JavaScript)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
  let left = 0;
  let right = nums.length - 1;
    
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    
    if (nums[mid] === target) {
      return mid;
    }
    
    // When dividing the roated array into two halves, one must be sorted.
    
    // Check if the left side is sorted
    if (nums[left] <= nums[mid]) {
      if (nums[left] <= target && target <= nums[mid]) {
        // target is in the left
        right = mid - 1;
        
      } else {
        // target is in the right
        left = mid + 1;
      }
    } 
    
    // Otherwise, the right side is sorted
    else {
      if (nums[mid] <= target && target <= nums[right]) {
        // target is in the right
        left = mid + 1;

      } else {
        // target is in the left
        right = mid - 1;
      }
    }
    
    
  }
    
  return -1;
};