Search in a Rotated Array [EASY]

Company:
MakemyTrip, Intuit, Hike, Flipkart, De-Shaw, BankBazaar, Amazon, Adobe, TimesInternet, Snapdeal, Paytm, Microsoft.

Question:

Given a sorted and rotated array (rotated at some point) A[ ], and given an element K, the task is to find the index of the given element K in the array A[ ]. The array has no duplicate elements. If the element does not exist in the array, print -1.
 

Input:
The first line of the input contains an integer T, depicting the total number of test cases. Then T test cases follow. Each test case consists of three lines. First line of each test case contains an integer N denoting the size of the given array. Second line of each test case contains N space separated integers denoting the elements of the array A[ ]. Third line of each test case contains an integer K denoting the element to be searched in the array.

Output:

Corresponding to each test case, print in a new line, the index of the element found in the array.  If element is not present, then print -1.

Constraints:

1 ≤ T ≤ 100
1 ≤ N ≤ 100005
0 ≤ A[i] ≤ 10000005
1 ≤ k ≤ 100005

Example:

Input
3
9
5 6 7 8 9 10 1 2 3
10
3
3 1 2
1
4
3 5 1 2
6

Output
5
1
-1

Code:

import java.util.*;

import java.lang.*;

import java.io.*;

class Solution {

public static void main (String[] args) {

//code

Scanner sc = new Scanner(System.in);

int m = sc.nextInt();

for(int i=0;i<m;i++){

   int n = sc.nextInt();

   int arr[]= new int [n];

   for (int j=0;j<n;j++){

       arr[j]=sc.nextInt();

       

   }

   int element = sc.nextInt();

   int count=0;

   for(int k=0;k<n;k++){

       if(arr[k]==element){

           System.out.println(k);

       }else {count++;}

   }

   if(count==n){System.out.println("-1");}

}

}

}