Fizz Buzz Java coding example

Q. Write a program that prints numbers from 1 to 30, but for multiples of three print "Fizz" instead of the number and for the multiples of five print "Buzz". For numbers which are multiples of both three and five print "FizzBuzz"?
A. This is a very basic coding question, but still some struggle or take too long (more than 5 minutes) to solve this coding problem. Be prepared for the interviewer to quiz you on different approaches to solve this coding problem. Be prepared to discuss DRY principle and optimization techniques to compare only once, etc.

There could be other variations. Arbitrary sets of number/word combinations (3 = Foo, 5=Bar, 7=Bazz, 11=Banana), etc.

One of the Java operators is "modulus", the name most programmers use for the remainder operator. It uses the symbol "%". Basically, "A % B" represents the remainder left over after dividing A by B. You use this operator to determine if a number is odd (i%2 != 0) or even (i%2 == 0).

Approach 1:

public class FizzBuzz {
 
    public static void main(String[] args) {
        StringBuilder builder = new StringBuilder(500);
        for (int i = 1; i <= 30; i++) {
            if (i % 15 == 0) {
               builder.append("FizzBuzz");
            }
            else if (i % 3 == 0) {
               builder.append("Fizz");
            }
            else if (i % 5 == 0){
              builder.append("Buzz");
            }
            else {
              builder.append(i);
            }
            builder.append('\n');
        }
        
        System.out.println(builder);
    }
}

Approach 2: Don't Repeat Yourself (DRY) approach with use of "Fizz" and "Buzz" not repeated. The boolean flag "processed" is used to determine if a number was multiples of 3, 5 or both.

public class FizzBuzzDry {
 
    public static void main(String[] args) {
        StringBuilder builder = new StringBuilder(500);
        for (int i = 1; i <= 30; i++) {
            boolean processed = false; 
            if (i % 3 == 0) {
               builder.append("Fizz");
               processed = true;
            }
            if (i % 5 == 0) {
               builder.append("Buzz");
               processed = true;
            }
   
            if (!processed) builder.append(i);
            builder.append('\n');
       }
        
       System.out.println(builder);
    }
}

Approach 3: Same as DRY approach shown above, but instead of using a boolean flag to determine if multiples of 3,5, or both, the length of the builder is compared before and after to see if already processed

public class FizzBuzzDry2 {

 public static void main(String[] args) {
  StringBuilder builder = new StringBuilder(500);
  for (int i = 1; i <= 30; i++) {
    int length = builder.length();
    if (i % 3 == 0)
      builder.append("Fizz");
    if (i % 5 == 0)
      builder.append("Buzz");
    if (length == builder.length())
      builder.append(i);
    
    builder.append('\n');
  }

  System.out.println(builder);
 }
}

I am sure there are many more approaches possible. For example, using a conditional operator, etc.

          builder.append(
          i % 15 == 0
                 ? "FizzBuzz"
                 : (i % 3 == 0
                     ? "Fizz"
                     : (i % 5 == 0
                         ? "Buzz"
                         : i)));
          builder.append("\n");