Java If-Else - Hacker Rank Solution

Using "if-else" you can perform decision making in Java. See the flowchart below (taken from wikipedia):



This problem will test your knowledge on "if-else" statements.
Given an integer N$N$ as input, check the following:

• If N$N$ is odd, print "Weird".
• If N$N$ is even and, in between the range of 2 and 5(inclusive), print "Not Weird".
• If N$N$ is even and, in between the range of 6 and 20(inclusive), print "Weird".
• If N$N$ is even and N>20$N>20$, print "Not Weird".

We given you partially completed code in the editor, complete it to solve the problem.

Input Format

There is a single line of input: integer N$N$.

Constraints

1≤N≤100$1\le N\le 100$

Output Format

Print "Weird" if the number is weird. Otherwise, print "Not Weird". Do not print the quotation marks.
Sample Input 1

3

Sample Output 1

Weird

Sample Input 2

24

Sample Output 2

Not Weird

 

Java If-Else - Hacker Rank Solution  

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        try{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int N=Integer.parseInt(br.readLine().trim());
        if(N<1||N>100)
            throw new Exception();
        if((N&1)==1)
        {
            System.out.println("Weird");
        }
        else
        {
            if(N>=2&&N<=5)
            {
                System.out.println("Not Weird");
            }
            else if(N>=6&&N<=20)
            {
                System.out.println("Weird");
            }
            else
            {
                System.out.println("Not Weird");
            }
        }
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }
    }
}

Java If-Else - Hacker Rank Solution