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Find the Second Occurrence of the Element

Get an array of Integers arr[] with size n and element e to be found. Apply Linear Search / Binary Search to find the element e’s second occurrence position in the array arr[]. If element e is found, print its second occurred index; if not found or occurred less than 2 times, print -1.

Examples

Input 1:

5

5 7 9 3 9

9

Output 1:

4

Input 2:

5

5 7 9 3 1

4

Output 2:

-1

Input 3:

5

5 7 9 3 1

9

Output 3:

-1

Try your Solution

Strongly recommended to Solve it on your own, Don't directly go to the solution given below.

public class Hello { public static void main(String args[]) { //Write your code here } }

Program or Solution

				
				
					
import java.util.*;
class SearchSO
{
	public static void main(String args[])
    	{

		//Variable and  Object Declarations
       		Scanner input = new Scanner(System.in);
		int n,e,pos=-1,count=0;
		
		//Getting size of Array & Declare Array arr[]
		n = input.nextInt();
		int arr[] = new int[n];

		//Get n values to array arr[]
		for(int i = 0; i<n; i++)
		{
			arr[i] = input.nextInt();
		}

		e = input.nextInt();

		//Linear Search
		for(int i = 0; i<n; i++)
		{
			if(arr[i]==e)
			{
				pos = i;
				count++;
				if(count == 2)
				{
					break;
				}
				
			}
		}
	
		if(count<2)
		{
			System.out.println(-1);
		}
		else
		{
			System.out.print(pos);
		}
		
      	}
}

Program Explanation

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