41. Search in Rotated Sorted Array

Topic :

binary search

Difficulty :

medium

Problem Link :


problem statement

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

Prior to being passed to your function, nums is possibly rotated at an unknown pivot 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 rotated at pivot index 3 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

import java.io.*;
import java.util.*;
class Search_An_Element_In_A_Sorted_and_Rotated_Array
{
  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 int search(int nums[],int key)
  {
      int low=0;
      int  high=nums.length-1;
      
      while(low<=high)
      {
          int mid=(low+high)/2;
          if(nums[mid]==key)
           return mid;
          
          //left half is sorted
          if(nums[low]<nums[mid])
         { 
          if(key>=nums[low] && key<nums[mid])
              high=mid-1; 
             else
              low=mid+1; 
            }
          // right half is sorted  
          else
          { if(key>nums[mid] && key<=nums[high])
              low=mid+1;
            else
             high=mid-1;
            }  
        }
      return -1;
    }
}
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