Java If-Else hackerrank solution

Java If-Else Hackerrank Solution

      
  import java.io.*;
  import java.math.*;
  import java.security.*;
  import java.text.*;
  import java.util.*;
  import java.util.concurrent.*;
  import java.util.function.*;
  import java.util.regex.*;
  import java.util.stream.*;
  import static java.util.stream.Collectors.joining;
  import static java.util.stream.Collectors.toList;
  
  public class Solution {
      public static void main(String[] args) throws IOException {
          BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
  
          int N = Integer.parseInt(bufferedReader.readLine().trim());
  
          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 {
                  System.out.println("Not Weird");
              }
          }
  
          bufferedReader.close();
      }
  }
  

Code Explanation

The code takes an integer input N from the user and uses if-else statements to perform the required conditional actions. Here's how the logic works:

If N is odd (N % 2 != 0), it prints "Weird."

If N is even (N % 2 == 0):

If N is between 2 and 5 (inclusive), it prints "Not Weird."

If N is between 6 and 20 (inclusive), it prints "Weird."

If N is greater than 20, it prints "Not Weird."

This logic fulfills the requirements of the given task.