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FizzBuzz Solution Dumping Ground


#266
for(int i=1;i<=100;i++)
{
            string val = string.Empty;
            if(i%3 == 0)
            {
                val = "Fizz";                
            }
            if(i%5 == 0)
            {
                val += "Buzz";
            }
            
            if(string.IsNullOrEmpty(val))
            {
                val = i.ToString();
            }
            Debug.Log(val);          
}

#267

some of these are cringe.

for (int i = 1; i <= 100; i++)
	{
		int mod3 = i%3, mod5 = i%5;
		if (!mod3 || !mod5)
		{
			if (!mod3) cout << "Fizz";
			if (!mod5) cout << "Buzz";
		}     		
		else cout << i;
     		cout << endl;
	}

#268

Here is the python version approach. This is my first actual working code since I started learning about couple of days

for x in range(1,22):
    #print x
    if x % 5 == 0:
        print x,"buzz"
    elif x % 3 ==0:
        print x,"fuzz"
    else:
        print x

Output:

SimpleTest.py’, wdir='C:/Users/rakz)
1
2
3 fuzz
4
5 buzz
6 fuzz
7
8
9 fuzz
10 buzz
11
12 fuzz
13
14
15 buzz
16
17
18 fuzz
19
20 buzz
21 fuzz


#269

F# Version with tail-recursion… Just because.

    let fizzBuzzList n =
        let rec fizzBuzz n acc =
            match n with
            | 0 -> acc
            | n when n % 3 = 0 && n % 5 = 0 -> fizzBuzz (n-1) ("FizzBuzz" :: acc)
            | n when n % 3 = 0 -> fizzBuzz (n-1) ("Fizz" :: acc)
            | n when n % 5 = 0 -> fizzBuzz (n-1) ("Buzz" :: acc)
            | _ as n -> fizzBuzz (n-1) (string n :: acc)
        fizzBuzz n []

    let printFizzBuzz n =
        fizzBuzzList n
        |> List.iter (fun s -> printfn "%s" s)

    printFizzBuzz 100000

#270

Recrusive Ruby Solution

def fizzbuzz(n)
  return if n > 100
  out = "#{n}: "
  out += "Fizz" if (n % 3 == 0)
  out += "Buzz" if (n % 5 == 0)
  puts out
  fizzbuzz(n+1)
end

fizzbuzz(1)

#271

It is really easy with Haskell and list comprehensions:

main :: IO ()
main = putStrLn $ unlines fizzbuzz

fizzbuzz :: [String]
fizzbuzz = [if x `dividable` 3
              then if x `dividable` 5
                      then "Fizzbuzz"
                      else "Fizz"
              else "Buzz"
            | x <- [1..100], x `dividable` 5 || x `dividable` 3]

dividable :: Int -> Int -> Bool
dividable lop rop = (lop `mod` rop) == 0

#272

A solution in F#

let (|Fizz|Buzz|FizzBuzz|Other|) x = 
    if x % 15 = 0 then FizzBuzz
    else if x % 5 = 0 then Fizz
    else if x % 3 = 0 then Buzz
    else Other

[1..100]
|> List.iter (fun x -> printfn "%s" (match x with
                                     | Fizz -> "Fizz"
                                     | Buzz -> "Buzz"
                                     | FizzBuzz -> "FizzBuzz"
                                     | _ -> x.ToString()))

#273
	for (i = 1; i <= 100; i++) {

		if (i%3 == 0 && i%5 == 0) {
			document.write(" - FizzBuzz <br />");
		} else if (i%3 == 0) {
			document.write(" - Fizz <br />");
		} else if (i%5 == 0) {
			document.write(" - Buzz <br />");
		} else {
			document.write(" - " + i + "<br />");
		}
	}

I am 12 and it took me a bit longer than 5 min because I wrote html.


#274

Oracle PL/SQL

DECLARE
   msg_ VARCHAR2(8);
BEGIN
   FOR rec_ IN (
      SELECT LEVEL AS lvl
        FROM DUAL
      CONNECT BY LEVEL <= 100
   ) LOOP
      SELECT DECODE(MOD(rec_.lvl, 15),
                    0, 'FizzBuzz',
                    3, 'Fizz',
                    5, 'Buzz',
                    6, 'Fizz',
                    9, 'Fizz',
                    10, 'Buzz',
                    TO_CHAR(rec_.lvl)) INTO msg_
        FROM DUAL;
      dbms_output.put_line(msg_);
  END LOOP;
END;
/

#275

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 :slight_smile:


#276

Swift…

for i in 1 ... 100 {
   let result = (i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "")
   print(result.isEmpty ? String(i) : result)
}

#277
#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;
}

#278

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])}


#280

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)])


#281
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();
        
    }
}