081. Search in Rotated Sorted Array II.md

Difficulty	Related Topics	Similar Questions
Medium	Array Binary Search https://leetcode.com/tag/array https://leetcode.com/tag/binary-search	Search in Rotated Sorted Array https://leetcode.com/problems/search-in-rotated-sorted-array

81. Search in Rotated Sorted Array II

Problem:

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

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

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

Solution:

See 33. Search in Rotated Sorted Array. The code is basically the same. Except with duplicates we can not tell which way to jump when pivot == nums[e]. The only thing we can do is to ditch nums[e]. SO worst case O(*n*).

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

  while (s <= e) {
    const p = (e + s) / 2 | 0
    const pivot = nums[p]

    if (target === pivot) {
      return true
    }

    if (pivot < nums[e]) {
      if (target > nums[p] && target <= nums[e]) {
        s = p + 1
      } else {
        e = p - 1
      }
    } else if (pivot > nums[e]) {
      if (target < nums[p] && target >= nums[s]) {
        e = p - 1
      } else {
        s = p + 1
      }
    } else {
      e--
    }
  }

  return false
};

Template generated via Leetmark.