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Fizz Buzz in Every Language

renegadecoder94 profile image Jeremy Grifski ・2 min read

Sample Programs Challenges (2 Part Series)

1) Fizz Buzz in Every Language 2) Fibonacci in Every Language

I noticed some folks on here were sharing some coding challenges, and I thought it might be fun to share some of my own (and shamelessly promote my Sample Programs project in the process).

The Challenge

Fizz Buzz is that notorious interview question almost everyone in the industry knows. Ironically, I don't know of anyone who has every been asked it. Regardless, it's a fun problem to tackle because it involves a little bit of everything: loops, conditions, math, etc. And, there are countless ways to solve it.

For the purposes of this challenge, here are the rules:

Write a program that prints the numbers 1 to 100. However, for multiples of three, print “Fizz” instead of the number. Meanwhile, for multiples of five, print “Buzz” instead of the number. For numbers which are multiples of both three and five, print “FizzBuzz”

Here's a segment of the output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

If you're feeling festive, you can swap the terms for whatever you like (i.e. EasterEgg, AprilShowers, etc.). In any case, the solution should be executable in any language of your choice. The more obscure the language, the better!

The Solution

I'll kick off the challenge with my favorite solution to the problem in Java:

public class FizzBuzz {
  public static void main(String[] args) {
    for (int i = 1; i < 101; i++) {
      String output = "";
      if (i % 3 == 0) {
        output += "Fizz";
      }
      if (i % 5 == 0) {
        output += "Buzz";
      }
      if (output.isEmpty()) {
        output += i;
      }
      System.out.println(output);
    }
  }
}

If you're curious how this solution works, Stuart Irwin wrote an excellent article about it. Along with this solution, I've been collecting several others in a repo called Sample Programs. After you drop your solution below, you should see if its missing from the collection. We'd love your contribution!

Sample Programs Challenges (2 Part Series)

1) Fizz Buzz in Every Language 2) Fibonacci in Every Language

Posted on by:

renegadecoder94 profile

Jeremy Grifski

@renegadecoder94

Engineering Education PhD student interested in challenging cultural issues in the tech community.

Discussion

markdown guide
 

There's several ways to do it in C++ :)

The Classic

#include <cstdio>
int main () {
  for (auto i = 1; i < 101; ++i) {
    if (i % 3 == 0) { std::printf("Fizz"); }
    if (i % 5 == 0) { std::printf("Buzz"); }
    std::printf(" %d\n", i);
  }
}

The "GOSH, MOM, IT'S GENERIC"

#include <cstdio>

struct number {
  number (int x) : x(x) { }
  auto operator * () const { return this->x; }
  bool operator == (int y) { return this->x == y; }
  bool operator != (int y) { return this->x != y; }
  number& operator ++ () { ++this->x; return *this; }
  number operator ++ (int) { return this->x++; }
  int x;
};


struct range {
    range (int start, int stop) :
      start(start),
      stop(stop)
    { }

    auto begin () const { return number(this->start); }
    auto end () const { return this->stop; }
private:
  int start;
  int stop;
};



int main () {
  for (auto i : range(1, 101)) {
    if (i % 3 == 0) { std::printf("Fizz"); }
    if (i % 5 == 0) { std::printf("Buzz"); }
    std::printf(" %d\n", i);
  }
}

The "Calculated At Compile Time (stare into madness edition)"

#include <type_traits>
#include <utility>
#include <cstdio>

template <char... Args>
struct as_string { const char data[sizeof... (Args)] = { Args... }; };

using fizz = as_string<'f', 'i', 'z', 'z'>;
using buzz = as_string<'b', 'u', 'z', 'z'>;
using newline = as_string<'\n'>;
using null = as_string<'\0'>;
struct fizzbuzz: fizz, buzz { };

template <class T> struct type_identity { using type = T; };

template <class T>
constexpr auto digits (T x) {
    auto digits = 0;
    while (x) {
        x /= 10;
        digits++;
    }
    return digits;
}

template <auto Size, auto X, char... Args>
struct to_string : to_string<Size - 1, X / 10, '0' + X % 10, Args...> { };

template <auto X, char... Args>
struct to_string<1, X, Args...> : type_identity<as_string<'0' + X, Args...>> { };

template <unsigned int X>
using int_to_string = typename to_string<digits(X), X>::type;

template <auto X, class Fizzable = std::bool_constant<X % 3 == 0>, class Buzzable = std::bool_constant<X % 5 == 0>> struct check;

template <auto X> struct check<X, std::false_type, std::false_type> : int_to_string<X> { };
template <auto X> struct check<X, std::true_type, std::false_type> : fizz, int_to_string<X> { };
template <auto X> struct check<X, std::false_type, std::true_type> : buzz, int_to_string<X> { };
template <auto X> struct check<X, std::true_type, std::true_type> : fizzbuzz, int_to_string<X> { };

template <auto X> struct calc : check<X>, newline { };
template <auto... X> struct print : calc<X>..., null { };

template <auto... Is>
print<(Is + 1)...> deduce (std::index_sequence<Is...>);

template <auto N>
using result = decltype(deduce(std::make_index_sequence<N>()));

static constexpr result<100> fb{};

int main() {
    std::puts(reinterpret_cast<const char *>(&fb));
    return 0;
}
 

I raise you the compile time calculated fizzbuzz in Nim ;)

proc fizzbuzz(): seq[string] =
  result = @[]
  for i in 1..100:
    result.add(
      if i mod 15 == 0: "FizzBuzz"
      elif i mod 5 == 0: "Buzz"
      elif i mod 3 == 0: "Fizz"
      else: $i
    )

const compileTimeValue = fizzbuzz()

echo compileTimeValue
 

C++20 isn't available yet, but we're a bit closer to this approach. :)

 
 

That last one scares me. Haha I’m not sure I’d know where to start reading it.

Thanks for jumping into the challenge with some great additions. 😃

 
 

number 3 is the Cthulu of FizzBuzz

 

CSS

ol {
   list-style-type: inside;
}

li:nth-child(3n), li:nth-child(5n) {
   list-style-type: none;
}

li:nth-child(3n):before {
   content: 'Fizz';
}

li:nth-child(5n):after {
   content: 'Buzz';
}
 

Haha I love that you can do this in CSS.

 

Here's my implementation in Racket

(define (mult-15? n)
  (and (mult-5? n)
       (mult-3? n)))

(define (mult-5? n)
  (= (modulo n 5) 0))

(define (mult-3? n)
  (= (modulo n 3) 0))

(define (fizzbuzz n)
  (cond
    [(mult-15? n) "FizzBuzz"]
    [(mult-5? n) "Buzz"]
    [(mult-3? n) "Fizz"]
    [else n]))

(define (print-list xs)
  (map displayln xs))

(print-list (map fizzbuzz (range 1 101)))

The print-list function is a bit redundant, since (map fizzbuzz (range 1 101)) will already print the resulting list to the console.

 

Ah come on @avalander - surely you should've written a FizzBuzz DSL in Racket? 😉

 

I should, but I don't know enough Racket for that yet 😅

 

Great stuff! I like racket a lot, but I haven't written any code in it myself. The fact that there are so many dialects of it is pretty cool to me.

 

Thanks! I've just started learning it myself. I can recommend the book Realm of Racket if you want to give it a shot.

 

Since nobody has done Javascript yet, here's a crazy implementation.

const worder = (predicate, patterner) => (prev, n) =>
  predicate(prev, n)
    ? patterner(prev, n)
    : prev

const isDivisible = d => (_, n) =>
  n % d === 0

const isEmpty = s =>
  s.length === 0

const append = x => (prev) => prev + x

const setNumber = (_, n) => n

const fizzer = worder(isDivisible(3), append('fizz'))

const buzzer = worder(isDivisible(5), append('buzz'))

const numberer = worder(isEmpty, setNumber)

const reducer = (...worders) => n =>
  worders.reduce(
    (prev, w) => w(prev, n),
    ''
  )

const fizzbuzzer = reducer(
  fizzer,
  buzzer,
  numberer
)

for (let i = 0; i <= 100; i++) {
  console.log(fizzbuzzer(i))
}

Consider how easy it is to extend to print 'fazz' for multiples of 7.

const fazzer = worder(isDivisible(7), append('fazz'))

const fizzbuzzfazzer = reducer(
  fizzer,
  buzzer,
  fazzer,
  numberer
)
 

I appreciate the commitment to the obscure. Haha these are great.

 

That's the whole point of the exercise, right? :D

Oh absolutely! Got any code golf solutions?

Hmm... the best I can come up with right now is 85 chars. Nothing really clever, just sacrificed readability for space.

let i=0;while(i++<101){console.log(i%15==0?'fizzbuzz':i%5==0?'buzz':i%3==0?'fizz':i)}

Another fun one, albeit longer, is this.

console.log(new Array(101)
  .fill(1)
  .map((_, i) =>
    i % 15 == 0 ? 'fizzbuzz' :
    i % 3 == 0 ? 'fizz' :
    i % 5 == 0 ? 'buzz' :
    i
  )
  .slice(1)
  .join('\n'))
 

Sql anybody?

DECLARE @i INT = 100;

;WITH E1(N) AS (SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
                SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1), -- 1*10^1 or 10 rows
      E2(N) AS (SELECT 1 FROM E1 a, E1 b),                                          -- 1*10^2 or 100 rows
      E4(N) AS (SELECT 1 FROM E2 a, E2 b),                                          -- 1*10^4 or 10,000 rows
      E8(N) AS (SELECT 1 FROM E4 a, E4 b),                                          -- 1*10^8 or 100,000,000 rows
cteTally(N) AS (SELECT TOP (@i) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E8)

SELECT  CASE WHEN N % 15 = 0 THEN 'fizzbuzz'
             WHEN N % 5 = 0 THEN 'buzz'
             WHEN N % 3 = 0 THEN 'fizz'  
             ELSE CONVERT(VARCHAR(10),N) END
FROM    cteTally


`

 

Because using a cursor is too mainstream? :)

 

Didn't see a C# solution, so here's mine (with some linq love):

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace FizzBuzz
{
    static class Program
    {
        static void Main(string[] args)
        {
            var buzzMap = new Dictionary<int, string>
            {
                { 3, "Fizz" },
                { 5, "Buzz" },
            };

            Func<int, string> buzzer = i =>
            {
                var buzzed = buzzMap.Keys
                    .Where(k => i % k == 0)
                    .Select(k => buzzMap[k]);

                return buzzed.Any() ? buzzed.Aggregate((prev, next) => prev += next) : null;
            };

            var output = Enumerable.Range(1, 100)
                .Select(i => buzzer(i) ?? i.ToString())
                .Aggregate(new StringBuilder(), (sb, s) => sb.AppendLine(s))
                .ToString();

            Console.WriteLine(output);
        }
    }
}
 

Elixir

1..100
|> Enum.map(fn
  n when rem(n, 3) == 0 and rem(n, 5) == 0 -> "FizzBuzz"
  n when rem(n, 3) == 0 -> "Fizz"
  n when rem(n, 5) == 0 -> "Buzz"
  n -> Integer.to_string(n)
end)
|> Enum.each(&IO.puts/1)
 

Here is another example using as a Module and using pattern matching. It looks hilarious :P

defmodule FizzBuzz do

  def run(num) when num >= 1 and num <= 100 do:
    fizzbuzz(num, rem(n, 3), rem(n, 5))
    run(num - 1)
  end

  def run(0), do: {:ok, "Done"}
  def run(_) do: {:error, "Something went wrong"}

  defp fizzbuzz(_, 0, 0), do: IO.puts "Fizzbuzz"
  defp fizzbuzz(_, 0, _), do: IO.puts "Fizz"
  defp fizzbuzz(_, _, 0), do: IO.puts "Buzz"
  defp fizzbuzz(n, _, _), do: n |> to_string |> IO.puts
end

# Example
FizzBuzz.run(55)
 
 

Here's some fun from the world of Common Lisp

Inoffensive version

(defun divides (divisor dividend)
  (= (mod divisor dividend) 0))

(defun fizzbuzz (n)
  (let ((fizz (when (divides n 3) "Fizz"))
        (buzz (when (divides n 5) "Buzz")))
    (if (or fizz buzz)
        (concatenate 'string fizz buzz)
        (write-to-string n))))

Offensive FizzBuzz Builder Macro

For when you want to define your own custom fizzbuzzer. Nice and easy to extend.

(defun divides (divisor dividend)
  (= (mod divisor dividend) 0))

(defmacro define-fizzbuzzer (name &body pairs)
  `(defun ,name (n)
     (let ((result nil))
       (dolist (p ',pairs)
         (let ((test-passed? (if (typep (first p) 'integer)
                                 (divides n (first p))
                                 (funcall (eval (first p)) n))))
           (when test-passed? (push (second p) result))))
       (if (null result)
           (write-to-string n)
           (apply #'concatenate 'string (nreverse result))))))

(define-fizzbuzzer fizz-buzz
  (3 "Fizz")
  (5 "Buzz"))

(define-fizzbuzzer fizz-buzz-bazz
  (3 "Fizz")
  (5 "Buzz")
  (7 "Bazz"))

Most Offensive Macrogeddon Too Hot for TV Version

For when you want to define your own custom matcher logic for each word in your custom fizzbuzzer. Still lets you put a single number in for 'divides by' logic.

(defmacro define-fizzbuzzer (name &body pairs)
  (let ((ps (clean-args pairs)))
    `(defun ,name (n)
       (let ((result nil))
         (dolist (p ',ps)
           (let ((test-passed? (funcall (eval (first p)) n)))
             (when test-passed? (push (second p) result))))
         (if (null result)
             (write-to-string n)
             (apply #'concatenate 'string (nreverse result)))))))

(defun clean-args (pairs)
  (mapcar #'clean-pair pairs))

(defun clean-pair (pair)
  (cond ((typep (first pair) 'integer)
         (list '#'(lambda (n) (divides n (first pair))) (second pair)))
        ((eq (caar pair) 'function)
         pair)
        (t (error "first element of pair must be either an integer or a function"))))

(defun prime? (n)
  "evaluates to t when n is prime. Highly inefficient"
  (loop for x from 2 to (round n 2) never (divides n x)))

(defun has-a-nine-in-it? (n)
 "does the decimal representation of this number have a 9 in it?"
 (some #'(lambda (c) (char= c #\9)) (write-to-string n)))

(define-fizzbuzzer fizz-buzz
  (3 "Fizz")
  (5 "Buzz"))

(define-fizzbuzzer fazz-bazz
  (#'prime? "Fazz")
  (#'has-a-nine-in-it? "Bazz"))
 
 
    print('\n'.join(['FizzBuzz' if x % 15 == 0 else 'Fizz' if x % 3 == 0 else 'Buzz' if x % 5 == 0 else str(x) for x in range(1,101)]))

 

Is this a Python solution using a massive list comprehension?! I like it.

 

I know this must be very basic, but it's the first time I'm ever doing this.
Here's my implementation using JavaScript, printed into an HTML element :D

<p id="target"></p>
let target = document.getElementById("target");

for(let i = 1; i <= 100; i++) {
  if( i % 5 == 0 && i % 3 == 0 ) {
    target.innerHTML += "FizzBuzz<br>";
  } else if( i % 5 == 0 ) {
    target.innerHTML += "Fizz<br>";
  } else if( i % 3 == 0) {
    target.innerHTML += "Buzz<br>";
  } else {
    target.innerHTML += i + "<br>";
  }
}
 

Here's a solution in Go.

package main

import (
    "fmt"
)

func main() {
    for num := 1; num < 101; num++ {
        if num % 3 != 0 && num % 5 != 0 {
            fmt.Println(num)
            continue
        }
        if num % 3 == 0 {
            fmt.Print("Fizz")
        }
        if num % 5 == 0 {
            fmt.Print("Buzz")
        }
        fmt.Println()
    }
}
 

Here's APL.

{(‘FizzBuzz’ ‘Fizz’ ‘Buzz’,⍵)[(0=15 3 5|⍵)⍳1]}¨⍳100

To understand how it works, read this

 

Love it! Short and sweet. I haven't gotten around to learning any APL, and the repo shows it.

 

Yep. I saw there wasn't an APL in the repo

 

Rather than post my own, I’m just going to link you to the fizzbuzz tag on Code Review. I swear I’ve reviewed this one in every language, and that includes LOLCODE.