FizzBuzz Solution Dumping Ground

practical · December 31, 2017, 8:59am

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])}

Rufus_V_Smith · February 3, 2018, 9:34pm

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)])

iGotTheGift · February 7, 2018, 5:36am

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

class FuzzBizzDriver
{
    private readonly IFizzBuzzRepository _fuzzBizzRepository;
    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);
        }
        Console.ReadKey();
    }
    private async Task<IList<string>> DoFizzBuzz(IEnumerable numbers)
    {
        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()
                }

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

Justin_Barrilleaux · February 22, 2018, 2:36am

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)
	}
}

gokumail · March 10, 2018, 2:39am

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");
}

Noura_Rizk · March 10, 2018, 2:39am

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);
            }
        }
    }

AttackWeasel · May 20, 2018, 9:06pm

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')

RSchiaffino · December 5, 2018, 12:01pm

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

fill series cells A1:A100 with numbers 1 through 100
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)))

Alan_Dev · January 12, 2019, 4:37pm

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++;
}

pero93 · January 31, 2019, 7:49pm

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);
        }

Jakub_Hronik · June 5, 2019, 5:13am

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.

Lee_Dydo · June 13, 2019, 6:50pm

Oh man. The article was right.

bitus · July 17, 2019, 2:59pm

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.

Umberto_Dos_Santos · August 30, 2019, 11:24am

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.

alrizopa · August 30, 2019, 11:24am

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";
}

alrizopa · August 30, 2019, 11:23am

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.

duds · August 30, 2019, 11:24am

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

JonatasAndrade · September 16, 2019, 2:03pm

Now imagine that poor soul that made Roller Coaster 2, ALL in assembly…

SomeRandomDude · October 12, 2019, 10:21pm

Let’s go python:

    def FizzBuzz(highestNumber, numbers, words):
	for i in range(1, highestNumber + 1):
		out = ""
		for number, word in zip(numbers, words):
			if i % number == 0:
				out += word
		print(out if out else i)

    FizzBuzz(100, [3,5], ["Fizz", "Buzz"])

Boom done. Reusable for different numbers and words

rvl · October 27, 2019, 9:48pm

Python, could be more compact but this way it is more clear what the program does and less computation only mod two times instead of four.

for i in range(0,101):
    
    fizz = i % 5 == 0
    buzz = i % 3 == 0
    
    if fizz and buzz:
        print("FizzBuzz")
        continue
    if fizz:
        print("Fizz")
        continue
    if buzz:
        print("Buzz")
        continue
    print(i)