Java Exception Handling using Try-catch

Java Exception Handling (Try-catch) HackerRank Problem and Solution 

Problem Description : 

You will be given two integers x and y as input, you have to compute x/y. If x and y are not 32 bit signed integers or if y is zero, exception will occur and you have to report it. Read sample Input/Output to know what to report in case of exceptions.

Learn about Exception and Exception handling :

• What is Exception and Exception handling in Java  
  

Java has built-in mechanism to handle exceptions. Using the try statement we can test a block of code for errors. The catch block contains the code that says what to do if exception occurs. 

Sample Input and Output :

Input 1

10 3

Output 1

3

Input 2

10 0

Output 2

java.lang.ArithmeticException: / by zero

Input 3

10 string

Output 3

java.util.InputMismatchException

Lets see solution : 

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
     
        try {
            Scanner in = new Scanner(System.in);
            int x = in .nextInt();
            int y = in .nextInt();
            System.out.println(x/y);
        } catch (ArithmeticException e) {
            System.out.println("java.lang.ArithmeticException: / by zero");
        } catch (InputMismatchException e) {
            System.out.println("java.util.InputMismatchException");
        }
    }
}

Solution explanation :

• Write all code in try block. so any Exception occur it will handled by catch() block.
  
• Get user input using Scanner class in x and y variable.
  
• Print solution of "x/y".
• If there will be divide by zero exception it will handled by "java.lang.ArithmeticException: / by zero". and if there will be input mismatch it will handled by "java.util.InputMismatchException" catch block.  

Happy coding...

See other hackerrank problem and solutions :

• Java Exception Handling HackerRank Soution with Explanation