FizzBuzz using limited number of conditions and StringBuilder

Question

All of you know the trivial fizzbuzz question during the first junior interviews. In my case, it was more than write a solution. There were more requirements:

1. Only two if\else statements.
2. StringBuilder required.
3. No Map, Collection.
4. No ternary operator.
5. Only Java core (8 or 11).
6. You have only 5 minutes (in my case, but for you it doesn't matter).

My solution with three if statements:

for (int i = 0; i <= 100; i++) {
    if (i%3==0 && i%5==0) {
        System.out.println("fizzBuzz");
    } else if (i%5==0) {
        System.out.println("Buzz");
    } else if (i%3==0) {
        System.out.println("fizz");
    } else {
        System.out.println(i);
    }
}

But yeah, there are three 'if' and no StringBuilder. Okay, let's see two 'if' example:

List<String> answer = new ArrayList<String>();
HashMap<Integer, String> map = new HashMap<Integer, String>();
map.put(3, "Fizz");
map.put(5, "Buzz");
for (int num = 1; num <= n; num++) {
    String s = "";
    for (Integer key : map.keySet()) {
        if (num % key == 0) {
            s += map.get(key);
        }
    }
    if (s.equals("")) {
        s += Integer.toString(num);
    }
    answer.add(s);
}

And again it's wrong: two 'if' but no StringBuilder.

Can I ask for a favor? Help me solve this problem.

Answer 1

I'd say your solution with the map was very close, and nobody said "no arrays" :)

public class FizzBuzz {

    public static void main(String args[]) {
        StringBuilder sb = new StringBuilder();
        Object[][] m = {{3, "Fizz"}, {5, "Buzz"}};
        for (int i = 1; i <= 100; i++) {
            boolean found = false;
            for (Object[] o : m) {
                if (i % (int) (o[0]) == 0) {
                    found = true;
                    sb.append((String) (o[1]));
                }
            }

            if (!found) {
                sb.append(i);
            }

            sb.append(" ");
        }

        System.out.println(sb.toString());
    }
}

Which prints

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz 28 29 FizzBuzz 31 32 Fizz 34 Buzz Fizz 37 38 Fizz Buzz 41 Fizz 43 44 FizzBuzz 46 47 Fizz 49 Buzz Fizz 52 53 Fizz Buzz 56 Fizz 58 59 FizzBuzz 61 62 Fizz 64 Buzz Fizz 67 68 Fizz Buzz 71 Fizz 73 74 FizzBuzz 76 77 Fizz 79 Buzz Fizz 82 83 Fizz Buzz 86 Fizz 88 89 FizzBuzz 91 92 Fizz 94 Buzz Fizz 97 98 Fizz Buzz

Having said that, I think 5 minutes is a bit too strict for this kind of problem.

Answer 2

Well, there's a simpler solution than the answer by Federico (in my humble opinion):

for (int i = 1; i <= 100; i++) {
    StringBuilder sb = new StringBuilder();
    if (i % 3 == 0) {
        sb.append("Fizz");
    }
    if (i % 5 == 0) {
        sb.append("Buzz");
    }
    String str = sb.toString();
    str = str.replaceAll("^$", String.valueOf(i));
    System.out.print(str + " ");
}

What happens here is that Fizz and Buzz are stored into the StringBuilder independently, so if the number is divisible by both 3 and 5, then the contents are FizzBuzz.

At last, we'll use a trick with replaceAll("^$", …) to conditionally replace the emtpy string by the number.

This actually yields

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz…

Instead of the trick with replaceAll, you could also use an Optional to simulate an if/else:

System.out.print(Optional.of(sb.toString())
    .filter(t -> !t.isEmpty())
    .orElse(String.valueOf(i)));

Answer 3

Try this.

for (int i = 1; i <= 100; i++) {
    StringBuilder sb = new StringBuilder();
    String number = Integer.toString(i);
    if (i % 3 == 0) {
        sb.append("Fizz");
        number = "";
    }
    if (i % 5 == 0) {
        sb.append("Buzz");
        number = "";
    }
    System.out.println(sb.append(number));
}

You can also do it without if-statement.

for (int i = 1; i <= 100; i++) {
    StringBuilder sb = new StringBuilder();
    boolean done = (i % 3 == 0 && sb.append("Fizz") == sb)
                 | (i % 5 == 0 && sb.append("Buzz") == sb)
                 || sb.append(i) == sb;
    System.out.println(sb);
}

Answer 4

import java.util.HashMap;
import java.util.Map;

public class FizzBuzz {

    public static void main(String[] args) {
        HashMap<Long, String> game = new HashMap<>(Map.of(
                3L, "Fizz",
                5L, "Buzz",
                23L, "Custom"));
        System.out.println(fizzBuzz(100, game));
    }

    public static String fizzBuzz(int size, HashMap<Long, String> game) {
        StringBuilder output = new StringBuilder();
        for (int i = 1; i <= size; i++) {

            boolean empty = true;
            for (Long divisor : game.keySet()) {
                if (i % divisor == 0) {
                    output.append(game.get(divisor));
                    empty = false;
                }
            }

            if (empty) {
                output.append(i);
            }
            output.append("\n");
        }

        return output.toString();
    }
}

Answer 5

import java.util.*;

public class FizzBuzz {

    public static void main(String[] args) {
        String[] params = {"Fizz\n3","Buzz\n5"};
        
        for (int i = 1; i ++< 100;) {
            StringBuilder buffer = new StringBuilder();
            for (String param: params) {
                Scanner reader = new Scanner(param);
                String str = reader.nextLine();
                int x = reader.nextInt();
                reader.close();
                if (i % x == 0) {
                    buffer.append(str);
                }
            }
            if (buffer.length() == 0) {
                buffer.append(i);
            }
            System.out.println(buffer.toString());
        }
    }

}