42. Search in Rotated Sorted Array II

Topic :

binary search

Difficulty :

medium

Problem Link :

problem statement

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before 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, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

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

Constraints:

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

solution

import java.io.*;
import java.util.*;
class Search_An_Element_In_A_Sorted_and_Rotated_Array_II
{
  public static void main (String args[])
  {
      int nums[]={4,5,6,7,0,1,2};
      int key=0;
      System.out.println(search(nums,key));
    }
    static boolean search(int[] nums, int target) {
        
        int low=0;
        int high=nums.length-1;
        
        while(low<=high)
        {
            int mid=low+(high-low)/2;
            
            if(nums[mid]==target)
                return true;
            if(nums[mid]==nums[low] && nums[mid]==nums[high]) // cannot determine if part of left sorted or right sorted
            {
                high--;
                low++;
            }
             else if(nums[mid]>=nums[low])// left sorted part
            {
                if(target>=nums[low] && target<nums[mid])
                    high=mid-1;
                else
                    low=mid+1;
            }
            else if (nums[mid]<nums[low])// right sorted part
            {
                if(target>nums[mid] && target<=nums[high])
                    low=mid+1;
                else
                    high=mid-1;
            }
            
        }
        
        return false;
    }
}