leetcode/work/1/Medium/Array/33.search-in-rotated-sorted-array.js

/*
 * 33. Search in Rotated Sorted Array
 * Difficulty: Medium
 * https://leetcode.com/problems/search-in-rotated-sorted-array/
 * 
 * ──────────────────────────────────────────────────
 * 
 * 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
 * 
 * 	• -10^4 <= nums[i] <= 10^4
 * 
 * 	• All values of nums are unique.
 * 
 * 	• nums is an ascending array that is possibly rotated.
 * 
 * 	• -10^4 <= target <= 10^4
*/

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

  while (l <= r){
    let mid = Math.floor((l + r)/2);
    if (target === nums[mid]) {
      return mid
    }

    // Left sorted portion
    if (nums[l] <= nums[mid]){
      if (target > nums[mid] || target < nums[l]){
        l = mid + 1;
      }else{
        r = mid - 1;
      }
    }
    // Right sorted portion
    else{
      if(target < nums[mid] || target > nums[r]){
        r = mid - 1; 
      }else{
        l = mid + 1;
      }
    }
  }
  return -1;
};