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;
}
}