1 ≤ N ≤ 100 you forgot this constraints if number is 101

The issue is with the string comparison using == instead of equals. == compares the references, which are very unlikely to be equal. Change to

if(grade.equals("A")){ totalGradePoints += 4; numClasses++; }else if(grade.equals("B")){ ... and it should work. See this answer for a detailed explanation.

As a good practice it is advisable to always use the static string as the object for calling equals on to prevent a NPE:

if("A".equals(grade)){ totalGradePoints += 4; numClasses++; }else if("B".equals(grade)){ ... If you are using Java 7, you can also do a switch statement with strings (though this one will throw an NPE if grade is null):

switch(grade) { case "A": totalGradePoints += 2; numClasses++; break; case "B": ... } And finally, since you are converting only one letter to an integer, to best solution is to convert them to char and for the values between A and D to do totalGradePoints += ('D' - grade.charAt(1)) + 1. So something along those lines would be simplest to read IMO:

while (true) {
final String input = in.next(); if(input == null || input.isEmpty()) break;

final char grade = input.charAt(0);
if(grade >= 'A' && grade <= 'D') {
    totalGradePoints += ('D' - grade) + 1;
} else if(grade == 'F') {
    // no gradepoints for 'F'
} else {
    break;
} //end if - else-if - else statement

++numClasses;

} //end while loop

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2nd one is very good and short logic appreciable!!!

Second Solution is smart :)

I think this code will not work currectly becouse user will insert whatever value then it will not check second condition becouse in case of OR if first condition is true then it will not check second condition. I hope you have understood

if(n%2==1 || (n>=6 && n <= 20))

Can you please explain actually what happens in this line ??? Why there is value 6 and 20 if our constraints are 1 to 100 ??

need explaination about on your logic can you do it please

ans = "Wierd"; if (n%2 == 0 && (n>20 || n < 6) { ans = "Not " + ans; }

for(int n=1;n<=100;n++) { do { if(n>=6 && n<=20) System.out.println("wierd"); else System.out.println("not wierd"); }while(n%2==0); System.out.println("wierd"); } //heyyy is this correct

its good place to code here

Hi,

Whilst your solution works, I am trying to figure out how you came to this solution? What was your way of thinking?

My solution is the following:

if (myInt%2==0 && (myInt >= 2 && myInt <= 5)){ return "Not Weird"; }else if (myInt%2==0 && (myInt >= 6 && myInt <= 20)){ return "Weird"; }else if (myInt%2==0 && myInt >20) { return "Not Weird"; }else{ return "Weird"; }

However, what I don't understand is that in your solution you're leaving out -> (myInt%2==0 && (myInt >= 2 && myInt <= 5) -> and it still works. Thanks. @inanutshellus

;) oh yes you do!

Good idea!

This is my favourite solution. Good stuff.

great (y)

You don't need parentheses - precedence of the ternary operator is lower than logical OR https://introcs.cs.princeton.edu/java/11precedence/

Wow... You're the real MVP.

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) {

        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();            
        String ans="";
        if(n%2==1){
          ans = "Weird";
        }
        elseif(n%2==0 && 2<=n<=5)
        {
        ans = "Not Weird";
        }
        elseif(n%2==0 && 6<=n<=20)
        {
            ans = "Weird";
        }
        elseif(n%2==0 && n>20)
        {
            ans = "Not Weird";
        }
        System.out.println(ans);

    }
}

It shows compilation error of semicolon after elseif statement.

U don't need the final IF

if(n>=2 && n<=5) ans = "Not Weird";

else if(n>=6 && n<=20) ans = "Weird";

else // ANY NUMBER GREATER THAN 20 ans = "Not Weird";

yes

yes. i too have got thi . even with 100 it shows the same

If n is even and in the inclusive range of 6 to 20, print "Weird"

That's one of the requirements.

does your code got submitted because mine code is runs successful bt at time of submit... it just shows loading symbol