 # FizzBuzz Solution Dumping Ground

REM before you judge, I am a senior level executive, not a programmer
REM I did some basic programming on Apple ii’s back in the 80’s
REM It took me about 10 mins, mostly shaking the rust off and reformatting some
REM of the earlier attempts

REM “FizzBuzz program”

for i = 1 to 100
if i mod 3 = 0 then REM i is evenly div by 3
print “Fizz”; REM suppress newline
end if
if i mod 5 = 0 then REM i is evenly div by 5
print “Buzz” REM with newline
else
print REM adds newline to "Fizz"
end if
REM we are done with div by 3, 5 and 15
if i mod 3 > 0 and i mod 5 > 0 then REM not div by 3 or 5
print i
end if
next i

REM I would fail most of the attempts listed here in an interview situation for failing to
REM properly document the code. Plain language documentation is important. Apologies
REM for not formatting the code. I didn’t know how Swift…

``````for i in 1 ... 100 {
let result = (i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "")
print(result.isEmpty ? String(i) : result)
}``````
``````#include <iostream>
using namespace std;

int main () {
for (int i = 1; i <= 100; ++i) {
if (i%3 == 0) {
cout << "Fizz";
}
if (i%5 == 0) {
cout << "Buzz";
}
if (i%3 != 0 && i%5 != 0) {
cout << i;
}
cout << "\n";
}
return 0;
}``````

`for(var i=1;i<101;++i){var c=0^((i%3)===0)|0^((i%5)===0)<<1; console.log([i,"Fizz","Buzz","FizzBuzz"][c])}`

Python 1 liner:
`print "\n".join([(str(i) if i%3 else 'Fizz') if i%5 else ("Buzz" if i%3 else "FizzBuzz") for i in range(1,101)])`

``````using System;
using System.IO;
using System.Collections.Linq;
using Pointless.Libs.Algorithms;

class FuzzBizzDriver
{
public FuzzBizzr()
{
_fizzBuzzRepository = ExcedinglyPointlessAbstractFactory.Create`<FizzBuzzRepository>`();
}

public static async void Main(string[] args...)
{
var numbers = Enumerable.Range(0,1111111);
var awsmResult = DoFizzBuzz(numbers);
for (var i = 0; i < numbers.Count; i++)
{
Console.WriteLine(awsmResult);
}
}
{
var unnecessaryDict = new Dictionary<int, string>();
Parallel.ForEach(
numbers,
new ParallelOptions { MaxDegreeOfParallelism = 16 },
number => {
var resultString = string.Empty;
resultString += _fizzBuzzRepository.Fizz(number, resultString);
resultString += _fizzBuzzRepository.Buzz(number, resultString);

if (string.IsNullOrWhiteSpace(resultString))
{
resultString = number.ToString()
}

});
return unnecessaryDict.OrderBy(x => x.Key).Select(x => x.Value).ToList();

}
}``````

Here is what i made in 5min and a little JS

``````for (var i = 1; i < 101; i++) {
if (i % 3 == 0 && i % 5 === 0) {
console.log("FizzBuzz")
}
else if (i % 3 === 0) {
console.log("Fizz")
}
else if (i % 5 === 0){
console.log("Buzz")
}
else{
console.log(i)
}
}
``````

My solution in C (in my defense, I didn’t write all this by hand ):

``````#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int main (){
printf("1 \n2 \nFizz \n4 \nBuzz \nFizz \n7 \n8 \nFizz \nBuzz \n11 \nFizz \n13 \n14 \nFizzBuzz \n16 \n17 \nFizz \n19 \nBuzz \nFizz \n22 \n23 \nFizz \nBuzz \n26 \nFizz \n28 \n29 \nFizzBuzz \n31 \n32 \nFizz \n34 \nBuzz \nFizz \n37 \n38 \nFizz \nBuzz \n41 \nFizz \n43 \n44 \nFizzBuzz \n46 \n47 \nFizz \n49 \nBuzz \nFizz \n52 \n53 \nFizz \nBuzz \n56 \nFizz \n58 \n59 \nFizzBuzz \n61 \n62 \nFizz \n64 \nBuzz \nFizz \n67 \n68 \nFizz \nBuzz \n71 \nFizz \n73 \n74 \nFizzBuzz \n76 \n77 \nFizz \n79 \nBuzz \nFizz \n82 \n83 \nFizz \nBuzz \n86 \nFizz \n88 \n89 \nFizzBuzz \n91 \n92 \nFizz \n94 \nBuzz \nFizz \n97 \n98 \nFizz \nBuzz \n");
}
``````
``````void loop(){
for(int i =1 ; i<=100; i++){
if (i% 15 == 0){
Log.i(TAG, "loop: FizzBuzz");
}else if(i % 3 == 0){
Log.i(TAG, "loop: Fizz");
}else if (i% 5 == 0){
Log.i(TAG, "loop: Buzz");
}else {
Log.i(TAG, "loop: " + i);
}
}
}
``````

Python 3:

``````print(*((x, 'Fizz', 'Buzz', 'FizzBuzz')[(not x%3) + (not x%5)*2] for x in range(1, 101)), sep='\n')
``````

or

``````print(*(x if x%5 and x%3 else 'FizzBuzz' if not x%3 and not x%5 else 'Fizz' if x%5 else 'Buzz' for x in range(1,101)), sep='\n')
``````

If nobody has done it yet in Excel (not VBA, simple Excel formulas):

1. fill series cells A1:A100 with numbers 1 through 100
2. fill down cells B1:B100 with formula `=IF(MOD(A1,15)=0,"FizzBuzz",IF(MOD(A1,5)=0,"Buzz",IF(MOD(A1,3)=0,"Fizz",A1)))`

Here’s a PHP solution that actually works ``````\$x=1;

while (\$x<=100)
{
if (\$x%3==0 and \$x%5==0)
{
print "<p>\tfizzbuzz</p>\n";
}
elseif (\$x%3==0)
{
print "<p>\tfizz</p>\n";
}
elseif (\$x%5==0)
{
print "<p>\tbuzz"."</p>\n";
}
else
{
print "<p>\$x</p>\n";
}

\$x++;
}``````

I see no continue being used.

``````for (int i = 1; i <= 100; i++)
{
if (i % 3 == 0 && i % 5 == 0)
{
Console.WriteLine("Fizz-Buzz {0}", i);
continue;
}
if (i % 3 == 0 && i % 5 != 0)
{
Console.WriteLine("Fizz {0}", i);
continue;
}
if (i % 3 != 0 && i % 5 == 0)
{
Console.WriteLine("Buzz {0}", i);
continue;
}
Console.WriteLine(i);
}``````

In C I would use this beautiful oneliner:

``````for(int t=1;t<=100;t++)
( (!(t%3)&&printf("Fizz")) + ((t%5==0)&&printf("Buzz")) || printf("%d", t)), putchar('\n');
``````

Just 2 modulos needed, no continue, no caching, just pure misuse of logical expression evaluation mechanics.

1 Like

Oh man. The article was right.

1 Like

One solution in Erlang with recursion.

``````-module(fizzbuzz).
-export([print/0]).

print() ->
FizzBuzzList = eval(100, []),
lists:foreach(fun(Item)->io:format("~p~n", [Item]) end, FizzBuzzList).

eval(Count, Acc) when Count /= 0 ->
Remainders = {Count rem 3, Count rem 5},
NewAcc = case Remainders of
{0, 0} -> ["FizzBuzz"|Acc];
{0, _} -> ["Fizz"|Acc];
{_, 0} -> ["Buzz"|Acc];
{_, _} -> [Count|Acc]
end,
eval(Count-1, NewAcc);

eval(0, Acc) ->
Acc.

``````
1 Like

Old but gold, vb6:

`````` Dim i As Integer
Open "C:\output.txt" For Append As #1 ' VB6 does not have native console window support
For i=1 To 100
If i Mod 3 = 0 Then Print #1, "Fizz"
If i Mod 5 = 0 Then Print #1, "Buzz"
If i Mod 15 <> 0 Then Print #1, Cstr(i)
Next
Close #1
``````

Also remembering that VB6 does not have an official console mode support, it’s necessary to waste two lines of code, openning and closing the text file used as Output.

1 Like

My 2 cents in Perl. Not the shortest, but it is among the most compact and still readable, doesn’t cheat using a module that does the work and is not exposed, and most importantly, it works CORRECTLY (version 5.10 and above due to use of “say” instead of “print”). Yes, it is amazing how many solutions here do not follow the actual specifications (or plain don’t work):

``````while (\$i++ < 100){
my \$out;
\$out  = "Fizz" unless (\$i%3);
\$out .= "Buzz" unless (\$i%5);
say "\$out" || "\$i";
}``````

This one is simpler, more readable, and no ternary ops:

``````while (\$i++ < 100){
my \$out;
\$out  = "Fizz" unless (\$i%3);
\$out .= "Buzz" unless (\$i%5);
say "\$out" || "\$i";
}

# ... and oh, it works correctly according to the specs.  One is not supposed to print the integer if the strings "fizz", "buzz", or "fizzbuzz" are printed (many solutions in this page disregard that point.``````

Of course we couldn’t forget our both loved and hated VBA in all its glory

Normal Loop:

``````Sub FizzBuzz(n As Integer)
Dim i As Integer

For i = 1 To n
If i Mod 3 = 0 And i Mod 5 = 0 Then
Debug.Print "FizzBuzz"
ElseIf i Mod 3 = 0 Then
Debug.Print "Fizz"
ElseIf i Mod 5 = 0 Then
Debug.Print "Buzz"
Else
Debug.Print i
End If
Next i

End Sub
``````

And some recursive:

``````Sub RecursiveFizzBuzz(n As Integer)

If n > 1 Then
RecursiveFizzBuzz (n - 1)
If n Mod 3 = 0 And n Mod 5 = 0 Then
Debug.Print "FizzBuzz"
ElseIf n Mod 3 = 0 Then
Debug.Print "Fizz"
ElseIf n Mod 5 = 0 Then
Debug.Print "Buzz"
Else
Debug.Print n
End If
Else
Debug.Print n
End If

End Sub
``````