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Java If Else – Hacker Rank Solution
In this challenge, we test your knowledge of using if-else conditional statements to automate decision-making processes. An if-else statement has the following logical flow:

Source: Wikipedia
Task
Given an integer , n , perform the following conditional actions:
- If n is odd, print
Weird
- If n is even and in the inclusive range of 2 to 5 , print
Not Weird
- If n is even and in the inclusive range of 6 to 20, print
Weird
- If n is even and greater than 20, print
Not Weird
Complete the stub code provided in your editor to print whether or not n is weird.
Input Format
A single line containing a positive integer, n .
Constraints
- 1<=n<=100
Output Format
Print Weird if the number is weird; otherwise, print Not Weird.
Example 0 :-
Sample Input 0
3
Sample Output 0
Weird
Example 1 :-
Sample Input 1
24
Sample Output1
Not Weird
Explanation
Sample Case 0: n=3n is odd and odd numbers are weird, so we print Weird
.Sample Case 1:n=24 n>24 and n is even, so it isn’t weird. Thus, we print Not Weird
.
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class Solution
{
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args)
{
int N = scanner.nextInt();
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
if(N%2!=0)
{
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 if(N>=20)
{
System.out.println("Not Weird");
}
scanner.close();
}
}
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