Writing FizzBuzz

Question

Reading the coding horror, I just came across the FizzBuzz another time.

The original post is here: Coding Horror: Why Can't Programmers.. Program?

For those who do not know: FizzBuzz is a quite popular children's game. Counting from 1 to 100, and every time a number is divisible by 3 the string "Fizz" is called, every time a number is divisible by 5 the string "Buzz" is called and every time a number is divisible by 3 and 5 both strings together "FizzBuzz" are called instead of the number.

This time, I wrote the code and it took me a minute, but there are several things that I do not like.

Here is my code:

public void DoFizzBuzz()
{
    var combinations = new Tuple<int, string>[] 
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    for (int i = 1; i <= 100; ++i)
    {
        bool found = false;

        foreach (var comb in combinations)
        {
            if (i % comb.Item1 == 0)
            {
                found = true;
                Console.Write(comb.Item2);
            }
        }

        if (!found)
        {
            Console.Write(i);
        }

        Console.Write(Environment.NewLine);
    }
}

So my questions are:

1. How do I get rid of the bool found?
2. Is there a better way of testing than the foreach?

Answer 1

I think your implementation is unnecessarily complex. This one does the job and is easier to understand:

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; i++)
    {
        bool fizz = i % 3 == 0;
        bool buzz = i % 5 == 0;
        if (fizz && buzz)
            Console.WriteLine ("FizzBuzz");
        else if (fizz)
            Console.WriteLine ("Fizz");
        else if (buzz)
            Console.WriteLine ("Buzz");
        else
            Console.WriteLine (i);
    }
}

Answer 2

Unrolled for maximum efficiency. This program can outfizzbuzz all others.

public void FizzBuzz()
{
    const string FIZZ = "Fizz";
    const string BUZZ = "Buzz";
    const string FIZZBUZZ = "FizzBuzz";

    int i = 0;
    while (i < 150)
    {
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(BUZZ); ++i;
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(BUZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(FIZZ); ++i;
        Console.WriteLine(++i);
        Console.WriteLine(++i);
        Console.WriteLine(FIZZBUZZ); ++i;
    }
}

Answer 3

Take advantage of conditional format specifiers to get a nicely golfed version:

public void DoFizzBuzz()
{
    for(int i=1;i<101;i++)Console.WriteLine("{0:#;}{1:;;Fizz}{2:;;Buzz}",i%3*i%5==0?0:i,i%3,i%5);
}

Answer 4

I think what you're trying to accomplish is a generic solution to FizzBuzz, that will work for any number of number-word combinations.

You have a good start - I think I can answer your questions with this example:

public void DoFizzBuzz()
{
    var combinations = new List<Tuple<int, string>>
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    Func<int, int, bool> isMatch = (i, comb) => i % comb == 0;
    for (int i = 1; i <= 100; i++)
    {
        Console.Write(i);

        var matchingCombs = combinations.Where(c => isMatch(i, c.Item1)).ToList();
        if (matchingCombs.Any())
        {
            Console.Write(string.Join("", matchingCombs.Select(c => c.Item2)));
        }
        else
        {
            Console.Write(i);
        }
        Console.Write(Environment.NewLine);
    }
}

In practice, you would pass combinations in to the method, but I included it inside just to be concise.

Answer 5

3rd edit:

Here is one way to "get rid of the bool" from your version (that is replace the for loop in your original question with this):

for (int i = 1; i <= 100; i++)
{
  var x = combinations.Where(n => i % n.Item1 == 0);

  if (x.Count() == 0)
    Console.Write(i);
  else
    Console.Write(string.Join("",x.Select(e => e.Item2)));

  Console.Write(Environment.NewLine);
}

---

Prior answers:

For a pure C# solution check out Keith Thompson's solution.

using System;
class FizzBuzz {
    static void Main() {
        for (int n = 1; n <= 100; n ++) {
            if (n % 15 == 0) {
                Console.WriteLine("FizzBuzz");
            }
            else if (n % 3 == 0) {
                Console.WriteLine("Fizz");
            }
            else if (n % 5 == 0) {
                Console.WriteLine("Buzz");
            }
            else {
                Console.WriteLine(n);
            }
        }
    }
}

I worked a bit on FixBuzz using linq. These are the solutions I came up with -- I believe they represent the best way to express the solution to this problem using Linq. (GitHub)

using System;
using System.Linq;

class FizzBuzz {
  static void Main() {
    var list = Enumerable.Range(1,100)
                .Select(n => {
                      if (n % 15 == 0) {
                        return "FizzBuzz";
                      }
                      if (n % 3 == 0) {
                        return "Fizz";
                      }
                      if (n % 5 == 0) {
                        return "Buzz";
                      }
                      return n.ToString();
                    });

    foreach(string item in list)
      Console.WriteLine(item);
  }
}

and the crazy one line version:

using System;
using System.Linq;

class FizzBuzz {
    static void Main() {
      Console.WriteLine(
      String.Join(
        Environment.NewLine,
        Enumerable.Range(1, 100)
          .Select(n => n % 15 == 0 ? "FizzBuzz" 
                     : n % 3 == 0 ? "Fizz" 
                     : n % 5 == 0 ? "Buzz" 
                     : n.ToString())
      ));
    }
}

Answer 6

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; i++)
    {
        if (i % 3 == 0)
            Console.Write("Fizz");
        if (i % 5 == 0)
            Console.Write("Buzz");
        if (!(i % 3 == 0 || i % 5 == 0))
            Console.Write(i);

        Console.Write(Environment.NewLine);
    }
}

This gets rid of the bool found, but forces you to do duplicate evaluation. It is slightly different from some of the other answers using i % 15 == 0 for the FizzBuzz qualification. Whether or not this is better is up for debate. However, it is a different way.

Answer 7

Did anyone do this one already?

Enumerable.Range(1, 100).Select(x =>
                (x % 15 == 0) ? "FIZZBUZZ"
                : (x % 5 == 0) ? "BUZZ"
                : (x % 3 == 0) ? "FIZZ"
                : x.ToString()
                )
                .ToList()
                .ForEach(console.WriteLine);

Answer 8

I think you started with a complicated way. Improving that code would be more complicated. You can use a temp variable to track and display that variable at the end of the FizzBuzz check. Below is code and you can also watch this detail c# FizzBuzz youtube video ( http://www.youtube.com/watch?v=OX5TM3q-JQg ) which explains how the below code is implemented.

    for (int j = 1; j <= 100; j++)
    {
    string Output = "";

    if (j % 3 == 0) Output = "Fizz";// Divisible by 3 --> Fizz

    if (j % 5 == 0) Output += "Buzz"; // Divisible by 5 --> Buzz

    if (Output == "") Output = j.ToString(); // If none then --> number

    Console.WriteLine(Output); // Finally print the complete output
    }

Answer 9

Will add my 5 cents to solution by Linq. Everybody is using Select, which is basically Map function. IMHO foldl function suits better to solve this quiz:

Console.WriteLine(
                Enumerable
                .Range(1, 100)
                .Aggregate(new StringBuilder(), (builder, i)
                    => i % 15 == 0 ? builder.AppendLine("FizzBuzz")
                     : i % 3 == 0 ? builder.AppendLine("Fizz")
                     : i % 5 == 0 ? builder.AppendLine("Buzz")
                     : builder.AppendLine(i.ToString()))
                .ToString());

Answer 10

Linq:

Enumerable.Range(1, 100).ToList().ForEach(i => Console.WriteLine( i % 3 * i % 5 == 0 ? (i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "") : i.ToString()));

Answer 11

In my opinion, the FizzBuzz problem is always presented as a challenge to the interviwee to make the word FizzBuzz appear without explicitly printing it. Here is my solution in C#.

internal void PrintFizzBuzzAlternative(int num)
{
    if (num % 5 == 0)
        Console.Write("Fizz");
    if (num % 3 == 0)
        Console.Write("Buzz");
    if (num % 5 != 0 && num % 3 != 0)
        Console.Write(num);
    Console.WriteLine();
}

Answer 12

Not the most efficient, but here's one using C#-6 string interpolation:

void Main()
{
    for (int i = 1; i <= 100; i++)
    {
       Console.WriteLine($"{(i % 15 == 0 ? "FizzBuzz" : 
                             i % 3 == 0 ? "Fizz" : 
                             i % 5 == 0 ? "Buzz" : i.ToString())}");
    }
}

Answer 13

Enumerable.Range(1, 100).ToList().ForEach(i=>Console.WriteLine($"{(i%3*i%5==0?0:i):#;}{i%3:;;Fizz}{i%5:;;Buzz}"));

This answer has it all:

• LINQ
• Conditional formatting
• String interpolation
• All on a single line

Victory!

Answer 14

The FizzBuzz question is a great interview question. We have started using it in our interview process. It is astounding how many people cannot solve such a simple problem.

Keep in mind, the original blog post was eventually locked due to a flood of people posting more solutions. Hahaha.

Regardless, here is mine in C++! ^_^

#include <iostream>
using namespace std;

int main(int argc, char** argv)
{
    for (int i = 1; i <= 100; ++i)
    {
        bool isMultipleOfThree = (i % 3) == 0;
        bool isMultipleOfFive = (i % 5) == 0;

        if (isMultipleOfThree) cout << "Fizz";
        if (isMultipleOfFive) cout << "Buzz";
        if (!isMultipleOfThree && !isMultipleOfFive) cout << i;

        cout << '\n';
    }

    return 0;
}

Answer 15

Ok, what the heck, here's the solution I've come to like :)

public void DoFizzBuzz()
{
    for (int i = 1; i <= 100; ++i)
    {
        bool isDivisibleByThree = i % 3 == 0;
        bool isDivisibleByFive = i % 5 == 0;

        if (isDivisibleByThree || isDivisibleByFive)
        {
            if (isDivisibleByThree)
                cout << "Fizz";

            if (isDivisibleByFive)
                cout << "Buzz";
        }
        else
        {
            cout << i;
        }
        cout << endl;
    }
}

Obviously, this is not the fastest solution, but I like it because it emphasizes readability and makes the "FizzBuzz" case no longer a special case, but something that will happen naturally through the code path.

In the end, what I love most about this question whenever it comes up is that we get to see just how many different solutions ppl can come up with.

Answer 16

I am a beginner, here is my attempt:

public void DoFizzBuzz()
   {
       for (int i = 1; i < 101; i++)
       {

           if ((i % 3 == 0) && (i % 5 == 0))
           {
               Console.WriteLine("{0} FizzBuzz", i);
           }
           else if (i % 3 == 0)
           {
               Console.WriteLine("{0} Fizz", i);
           }
           else if (i % 5 == 0)
           {
               Console.WriteLine("{0} Buzz", i);
           }
           else
           {
               Console.WriteLine(i);
           }

       }
       Console.ReadLine();
   }

Is there anything wrong with my approach? Mine seems a lot simpler than everyone's else approach so it must be wrong.

Answer 17

You want probably make it configurable, but the question is what should be made configurable - we don't know that. Maybe we should make configurable all the cycle (FizzBuzz has the cycle). Here is very small and fun version with configurable cycle:

string[] fizzBuzzCycle = 
    "FizzBuzz,{0},{0},Fizz,{0},Buzz,Fizz,{0},{0},Fizz,Buzz,{0},Fizz,{0},{0}"
    .Split(',');

for (int i = 1; i <= 100; i++)
    Console.WriteLine(fizzBuzzCycle[i%fizzBuzzCycle.Length], i);

So if the strings or whole cycle should be changed it is easy to change. But you just don't know what to make configurable. Maybe condition will change: "for prime numbers print Pizz" and for this modification the solution by @ThomasLevesque is better, because it is easier to change.

Answer 18

I tried to solve this problem without looking at the answers. It took me 3 hours to succeed. (I'm just a hobby programmer by the way so don't bash me hard please :)) This is my c# version solution:

        static void Main(string[] args)
    {

        for (int i = 1; i <= 100; i++)
        {
            if(  ((i % 3) != 0) && ((i % 5) != 0))
            {
                WriteLine($"{i}");
            }
            else
            {
                if ((i % 15) == 0)
                {
                    WriteLine("FizzBuzz");
                }
                else if ((i % 3) == 0)
                {
                    WriteLine("Fizz");
                }
                else if ((i % 5) == 0)
                {
                    WriteLine("Buzz");
                }
            }                 
        }
    }

Answer 19

With no if conditions, just one ternary operator.

string[] s = new string[6]{"Fizz", "Buzz", "", "", "", ""};
for (int i = 1; i <= 100; i++)
{
    string output = s[(i%3)*2] + s[(i%5)+1];
    Console.WriteLine(string.IsNullOrEmpty(output)? "" + i : output);
}

Answer 20

The null-coalescing operator is really useful:

string output = null;
for (int i = 1; i <= 100; i++)
{
     if (i % 3 == 0) output += "fizz";
     if (i % 5 == 0) output += "buzz";
     Console.WriteLine(output ?? i.ToString());
     output = null;
}
Console.ReadKey();

Answer 21

I recommend using the ++i instead of the i++ in a for loop because i++ requires a copy to be made ;)

public void DoFizzBuzz()
{
    for (int i = 1; i < 101; ++i)
    {
        if (i % 15 == 0)
            Console.WriteLine ("FizzBuzz");
        else if (i % 3 == 0)
            Console.WriteLine ("Fizz");
        else if (i % 5 == 0)
            Console.WriteLine ("Buzz");
        else
            Console.WriteLine (i);
    }
}

Answer 22

With the input of Rob H and Jacob Krall here is what I have at the moment. Perhaps I will play around with that in future... just wanted to provide it.

public void DoFizzBuzz()
{
    // expect this to come in as parameter
    var combinations = new Tuple<int, string>[] 
    { 
        new Tuple<int, string> (3, "Fizz"), 
        new Tuple<int, string> (5, "Buzz"), 
    };

    Func<int, int, bool> isMatch = (i, comb) => i % comb == 0;

    // expect the borders 1, 100 to come in as parameters
    for (int i = 1; i <= 100; ++i)
    {
        var matchingCombs = combinations.Where(c => isMatch(i, c.Item1)).DefaultIfEmpty(new Tuple<int, string>(i, i.ToString())).Aggregate((v, w) => new Tuple<int, string>(v.Item1, v.Item2 + w.Item2)).Item2;
        Console.WriteLine(matchingCombs);
    }
}

Answer 23

I would suggest this compact code as an addition to the previous simple and nice versions.

for (int i = 1; i <= 100; i++) // i++ but not ++i as in your example, be careful here
{
    bool fizz = i % 3 == 0;
    bool buzz = i % 5 == 0;
    string output = fizz && buzz ? "FizzBuzz" :
                            fizz ? "Fizz" :
                            buzz ? "Buzz" :
                            i.ToString();

    Console.WriteLn(output);
}

Answer 24

A functional approach...

Console.WriteLine(Enumerable
    .Range(1,100)
    .Aggregate("", 
        (a,i) => a + "\n" + (i%15==0 ? "fizzbuzz" : 
                                (i%5==0 ? "buzz" :
                                    (i%3==0 ? "fizz" : i.ToString())))));

Answer 25

Relatively simple solution using a for loop.

No Linq or anything - just basic shorthand if statements

for(int x=1;x<101;x++)
    Console.WriteLine(x%3==0?"Fizz"+(x%5==0?"Buzz":""):x%5==0?"Buzz":x+"");

The Linq solution which is a lot like csells (sans string interpolation) and fits on one line would be:

Enumerable.Range(1,100).ToList().ForEach(x=>Console.WriteLine(x%3==0?"Fizz"+(x%5==0?"Buzz":""):x%5==0?"Buzz":x+""));

Answer 26

Obviously this is a bit outside the spirit of the FizzBuzz challenge. But in my benchmark this was the fastest I could make it while single threaded and still terminating at 100. It is semi-unrolled and uses a StringBuilder. It is approximately three times faster than the standard approach.

const string FIZZ = " Fizz\n";
const string BUZZ = " Buzz\n";
const string FIZZBUZZ = " FizzBuzz\n";
    ...
var sb = new StringBuilder();
int i = 0;

while(true)
{       
    sb.Append(i+3);
    sb.Append(FIZZ);        
    sb.Append(i+5);
    sb.Append(BUZZ);        
    sb.Append(i+6);
    sb.Append(FIZZ);        
    sb.Append(i+9);
    sb.Append(FIZZ);        
    sb.Append(i+10);
    sb.Append(BUZZ);        
    if(i+12 > 100)
        break;
    sb.Append(i+12);
    sb.Append(FIZZ);    
    i+=15;
    sb.Append(i);
    sb.Append(FIZZBUZZ);
}

Console.Write(sb.ToString());

Answer 27

I'll add mine even though there's 20 other solutions already written: It goes like this....

var x = 1;
while (x <= 100)
{
     if (x % 3 == 0 && x % 5 == 0)
        {Console.Writeline("FizzBuzz");}
     else if (x % 3 == 0)
        {Console.Writeline("fizz");}
     else if (x % 5 == 0)
        {Console.Writeline("Buzz");}
     else
        {Console.Writeline(x);}
     x++ 
}

First solution I came up with. Simple, to the point and gets the job done. No need for bool.

Answer 28

This my effort mixing Func with IEnumerable

 class Program
{
    static void Main(string[] args)
    {
        foreach (var i in FizzBuzz(100))
        {
            Console.WriteLine(i);
        }
    }

    private static IEnumerable<string> FizzBuzz(int maxvalue)
    {
        int count = 0;
        //yield return count.ToString();
        Func<int, string> FizzBuzz = (x) => ((x % 5 == 0 && x % 3 == 0) ? "FizzBuzz" : null);
        Func<int, string> Buzz = (x) => ((x % 5 == 0) ? "Buzz" : null);
        Func<int, string> Fizz = (x) => ((x % 3 == 0) ? "Fizz" : null);
        Func<int, string> Number = (x) => x.ToString();

        while (count < maxvalue)
        {
            count++;

            yield return FizzBuzz(count) ?? Buzz(count) ?? Fizz(count) ?? Number(count);
        }
    }
}

Answer 29

The original questions were: 1.How to get rid of the bool found? 2.Is there a better way of testing than the foreach?

This gets rid of the bool and the foreach, and I think it's still readable.

public static void DoFizzBuzz()
{
    var combinations = new Tuple<int, string>[]  
    {  
        new Tuple<int, string> (3, "Fizz"),  
        new Tuple<int, string> (5, "Buzz"),  
    };

    for (int i = 1; i <= 100; i++)
    {
        var fb = combinations.Where(t => {
            if (i % t.Item1 == 0)
            {
                Console.Write(t.Item2);
                return true;
            }
            return false;
        }).ToList();

        if (!fb.Any())
        {
            Console.Write(i);
        }

        Console.Write(Environment.NewLine);
    }
} 

Who'd a thunk we'd be getting so excited about a simple kids game? :)

Answer 30

You can use either use this and only take the amount you want

static void Main(string[] args)
{
    GetFizzBuzz().Take(100).ToList().ForEach(Console.WriteLine);
}

private static IEnumerable<string> GetFizzBuzz()
{
    for (var i = 0; i < int.MaxValue; i++)
    {
        if (i % 3 == 0 && i % 5 == 0) yield return "FizzBuzz";
        if (i % 3 == 0) yield return "Fizz";
        yield return i % 5 == 0 ? "Buzz" : i.ToString(CultureInfo.InvariantCulture);
    }
}

Or simply use this :

Enumerable.Range(1, 100).Select(s => {
    if (s % 3 == 0 && s % 5 == 0) return "FizzBuzz";
    if (s % 3 == 0) return "Fizz";
    return s%5 == 0 ? "Buzz" : s.ToString(CultureInfo.InvariantCulture);
}).ToList().ForEach(Console.WriteLine);

Answer 31

Found Stringbuilder to be the fastest.

I find this implementation the easiest to read as well.

public static void FizzBuzz() {

    StringBuilder sb = new StringBuilder();

    for (int i = 1; i <= 100; i++) {    
        if (i % 3 == 0 && i % 5 == 0) {
            sb.AppendLine($"{i} - FizzBuzz");
        } else if (i % 3 == 0) {
            sb.AppendLine($"{i} - Fizz");
        } else if (i % 5 == 0) {
            sb.AppendLine($"{i} - Buzz");
        } else {
            sb.AppendLine($"{i}");
        }
    }
    Console.WriteLine(sb);
}

Answer 32

Without using any If's, C# code.

 //False = 0, True = 1.
    private void DivisibilityByFiveThreeTest(int num)
    {
        string[,] values = new string [2,2]{
                             {"None","Fizz"},
                             {"Buzz","FizzBuzz"}
                             };
        for(int i=1;i< num;i++)
        Console.WriteLine(values[Convert.ToInt32(i % 5 == 0), Convert.ToInt32(i%3==0)]);

    }

Answer 33

straight forward solution in JavaScript

var i = 1;
while (i <= 100){
    console.log((i % 3 === 0 && i % 5 === 0) ? "FizzBuzz" : (i % 3 === 0) ? "Fizz" : (i % 5 === 0 ? "Buzz" : i));
    i++;
}

Answer 34

In python.....

 for i in range(0, 100) :
        a = i % 3 == 0
        b = i % 5 == 0

        if(a & b):
            print("FizzBuzz")
        elif(a):
            print("Fizz")
        elif(b):
            print("Buzz")
        else:
            print(i)

Answer 35

C++ implementation

vector<string> FizzBuzz::execute(int n) {

    if(n == 0)
        return vector<string>();

    auto push = n % 15 == 0 ? "fizzbuzz" :
                n % 3 == 0 ? "fizz" :
                n % 5 == 0 ? "buzz" :
                to_string(n);

    auto execution = execute(n-1);
    execution.push_back(push);
    return execution;
}

Answer 36

In C#...

using System;
using System.IO;
class Solution {
    static void Main(String[] args) {        
        for(int i=1;i<=100;i++)
        {                       
         string result=(i%3==0 && i%5==0) ? "FizzBuzz" : 
               (i%5==0) ? "Buzz" :
                       (i%3==0) ? "Fizz" : i.ToString();            
         Console.WriteLine(res);
        }
    }
}


In VB.NET...

Imports System
Imports System.IO

Class Solution
    Private Shared Sub Main(ByVal args As String())
        For i As Integer = 1 To 100
            Dim res As String = If((i Mod 3 = 0 AndAlso i Mod 5 = 0), "FizzBuzz", If((i Mod 5 = 0), "Buzz", If((i Mod 3 = 0), "Fizz", i.ToString())))
            Console.WriteLine(res)
        Next
    End Sub
End Class

Answer 37

Been answered to death but just to show another solution that answers the actual question

public static void DoFizzBuzz()
{
    var combinations = new (int multiple, string output)[]
    {
        (3, "Fizz"),
        (5, "Buzz")
    };

    for (int i = 1; i <= 100; ++i)
    {
        // Seed the accumulation function with an empty string and add any 'matches' from each combination
        var fb = combinations.Aggregate("", (c, comb) => c + (i % comb.multiple == 0 ? comb.output : ""));

        Console.WriteLine(!string.IsNullOrEmpty(fb) ? fb : $"{i}");
    }
}

Answer 38

This is an answer written in JavaScript

//returns true if modulus is zero and is reusable        
var isModulusZero = (challenge, modulator) => {
    let answer = challenge % modulator;

    if (answer == 0) {
        return true
    } else {
        return false
    }
}

var printOneToHundred = () => {
    var i = 1;

    for (i; i <= 100; i++) {

        if (isModulusZero(i, 3) && isModulusZero(i, 5)) {
            console.log('FizzBuzz')
        } else if (isModulusZero(i, 3)) {
            console.log('Fizz')
        } else if (isModulusZero(i, 5)) {
            console.log('Buzz')
        } else {
            console.log(i)
        }

    }
}

Answer 39

Ok - I'll bite. The LINQ heavy solution I came up with on a recent initial coding interview. Amazing how many different solutions there are to this simple problem.

public static class FizzBuzzUtils
{
    private static List<KeyValuePair<int, string>> map = new List<KeyValuePair<int, string>> {
        new KeyValuePair<int, string>(3, "Fizz"),
        new KeyValuePair<int, string>(5, "Buzz")
    };

    public static string GetValue(int i)
    {
        var matches = map.Where(kvp => i % kvp.Key == 0)
            .Select(kvp => kvp.Value)
            .ToArray();
        return matches.Length > 0 ? string.Join(string.Empty, matches) : i.ToString();
    }

    public static IEnumerable<string> Range(int start, int count)
    {
        return Enumerable.Range(start, count)
            .Select(i => GetValue(i));
    }
}

Answer 40

Fizz Buzz Pattern (see CachedFizzBuzz method), combined with strategy pattern.

using System;
interface IFizzBuzz
{
}
class NullFizzBuzz: IFizzBuzz
{
    public static int value;
     public override string ToString()
    {
        return value.ToString();
    }
}
class Buzz : IFizzBuzz
{
    public override string ToString()
    {
        return "Buzz";
    }
}
class Fizz: IFizzBuzz
{
    public override string ToString()
    {
        return "Fizz";
    }
}
class FizzBuzz : IFizzBuzz
{
    public override string ToString()
    {
        return "FizzBuzz";
    }
}

class FizzBuzzSolver
{
    public int Fizz { get; }
    public int Buzz { get; }
    public int Iterations { get; }
    public int FizzBuzz { get; }
    public int Quotient { get; }
    public int Remainder { get; }
    public FizzBuzzSolver(int fizz, int buzz, int iterations, int fizzbuzz, int quotient, int remainder)
    {
        Fizz = fizz;
        Buzz = buzz;
        Iterations = iterations;
        FizzBuzz = fizzbuzz;
        Quotient = quotient;
        Remainder = remainder;
    }

    internal void Solve(Action<FizzBuzzSolver> fizzBuzz)
    {
        fizzBuzz(this);
    }
}
class Program
{
    static void Main()
    {
        int fizz = 3;// int.Parse(Console.ReadLine());
        int buzz = 5;// int.Parse(Console.ReadLine());
        int iterations = 100;//int.Parse(Console.ReadLine());
        int fizzbuzz = fizz * buzz;
        int quotient = Math.DivRem(iterations, fizzbuzz, out var remainder);
        FizzBuzzSolver fb = new FizzBuzzSolver(fizz, buzz, iterations, fizzbuzz, quotient, remainder);
        switch (quotient)
        {
            case 0: fb.Solve(RemainderFizzBuzz); break;
            case 1: fb.Solve(NormalFizzBuzz);  break;
            default: fb.Solve(CachedFizzBuzz); break;
        };

        Console.Read();
    }

    private static void NormalFizzBuzz(FizzBuzzSolver fb)
    {
        for (int i = 1; i < fb.FizzBuzz; i++)
        {
            Console.WriteLine(i % fb.Fizz == 0 ? "Fizz" : i % fb.Buzz == 0 ? "Buzz" : i.ToString());
        }
        Console.WriteLine("FizzBuzz");
        for (int i = fb.FizzBuzz+1; i <= fb.Iterations; i++)
        {
            Console.WriteLine(i % fb.Fizz == 0 ? "Fizz" : i % fb.Buzz == 0 ? "Buzz" : i.ToString());
        }
    }

    private static void CachedFizzBuzz(FizzBuzzSolver fb)
    {
        var fizzbuzzArray = new IFizzBuzz[fb.FizzBuzz];

        for (int i = 0; i < fb.FizzBuzz - 1; i++)
        {
            int s = i + 1;
            if (s % fb.Fizz == 0)
            {
                fizzbuzzArray[i] = new Fizz();
                Console.WriteLine("Fizz");
            }
            else
            if (s % fb.Buzz == 0)
            {
                fizzbuzzArray[i] = new Buzz();
                Console.WriteLine("Buzz");
            }
            else
            {
                fizzbuzzArray[i] = new NullFizzBuzz();
                Console.WriteLine(s);
            }

        }
        fizzbuzzArray[fb.FizzBuzz - 1] = new FizzBuzz();
        Console.WriteLine("FizzBuzz");
        NullFizzBuzz.value = fb.FizzBuzz;
        for (int j = 1; j < fb.Quotient; j++)
        {
            for (int i = 0; i < fb.FizzBuzz; i++)
            {
                NullFizzBuzz.value++;
                Console.WriteLine(fizzbuzzArray[i]);
            }

        }
        for (int i = 0; i < fb.Remainder; i++)
        {
            NullFizzBuzz.value++;
            Console.WriteLine(fizzbuzzArray[i]);
        }
    }

    private static void RemainderFizzBuzz(FizzBuzzSolver fb)
    {
        for (int i = 1; i <= fb.Remainder; i++)
        {
            Console.WriteLine(i % fb.Fizz == 0 ? "Fizz" : i % fb.Buzz == 0 ? "Buzz" : i.ToString());
        }
    }
}

Answer 41

Unlike other answers here, this solution offers the possibility of easily extending it. If an interviewer were to ask you to e.g. also replace 7 with "Woof", it is easily done by adding to the map.

var map = new Dictionary<int, string>
{
    {3*5*7, "FizzBuzzWoof"}, // Optional
    {5*7, "BuzzWoof"}, // Optional
    {3*7, "FizzWoof"}, // Optional
    {3*5, "FizzBuzz"},
    {7, "Woof"},
    {5, "Buzz"},
    {3, "Fizz"},
    {1, null} // For all numbers not evenly divisible
};

var words = Enumerable
    .Range(0, 154)
    .Select(i => (i == 0 ? null : map[map.Keys.First(k => i % k == 0)]) ?? i.ToString())
    .Dump("Words");

string.Join(' ', words).Dump("Result");

Identifying all permutations, however, becomes a chore, and the more extra words are added, the more lookups will be made, but the point is to argue about understanding coding principles.

Answer 42

Since we are going for a generic FizzBuzz (I've seen this called Raindrops), there are a couple simple modifications to keep it generic and more readable:

1. Renaming 'combinations' to 'primeFactors' makes it a little more descriptive
2. Tuple field names to make accessing tuple fields more readable
3. Linq to query the list of prime factors for the display strings we want (removing the foreach)
   • Where filters the list to factors of the current integer
   • Select transforms the matching factors into their display string
   • DefaultIfEmpty handles the 'no match found' case allowing us to remove that boolean
   • Aggregate concatenates all the matching display strings together

Try it online!

public void DoFizzBuzz()
{
    var primeFactors = new (int Factor, string Display)[]
    {
        (3, "Fizz"),
        (5, "Buzz")
    };
    for (var integer = 1; integer <= 100; ++integer)
        Console.WriteLine(primeFactors
            .Where(tuple => integer % tuple.Factor == 0)
            .Select(tuple => tuple.Display)
            .DefaultIfEmpty($"{integer}")
            .Aggregate((first, second) => $"{first}{second}")
        );
}

Answer 43

var min = 1;
var max = 100;

for (var i = min; i <= max; i++)
{
    if (i % 3 == 0) { goto fizz; }

checkBuzz:
    if (i % 5 == 0) { goto buzz; }
    goto writeNumber;

fizz:
    Console.Write("Fizz");
    goto checkBuzz;

buzz:
    Console.Write("Buzz");
    goto allDone;

writeNumber:
    if (i % 3 != 0) { Console.Write(i); }

allDone:
    Console.WriteLine();
}

Answer 44

My final approach to FizzBuzz in C#.

There are countless solutions to the FizzBuzz test but few meet my requirements in crisp clear, readable, clean and maintainable code. Let me just say what I mean with

1. clear: isolate a task into kind of logic atoms and encapsulate them inside a type
2. readable: think of identifier names that describe and fit to the task; format your code
3. clean: design the members of the type with as little “noise” as possible
4. maintainable: maybe you want to or have to extend the type in the future. think of a layout that will allow that

Here is the solution I came up with. A static class that handles the task.

My first question was: what are the possible states of FizzBuzz? Ok, either FIZZ or BUZZ, that's clear. And there is also the FIZZBUZZ state if the number can be divided by 3 and 5 at the same time. Hm, interesting. This can be regarded as a (logical) sum of FIZZ + BUZZ… Finally, the state NONE (or 0) remains if neither state fits.

As an experienced developer with old low level skills (asm), I decided to use a so-called bit vector (binary flags) to fit my task. The .NET framework also makes heavy use of these when it comes to configuration options and offers a comfortable C# language construct, the enum type. Due to public visibility, I name this enum Category to depict the possible states.

Next there are the intern FIZZ and the BUZZ functions that perform the divisibility test on the given number and return the appropriate category.

Finally the only publicly available function named “test” allows us to test a given number. It returns the number’s state as defined in Category. The trick here is, that the function calls both members fizz and buzz and combines or adds (ok, boolean ORs) their results to the proper state.

I was also experimenting with some functional approaches using lambdas but these had no advantages over the solution shown. I like to keep things simple.

Happy coding,

Tom

Download Gist: https://gist.github.com/tomschrot/da0bd69dd1e591fa3da43c1fdb3e4f85

WriteLine ("Final FizzBuzz in C#\nby Tom Schröter\n");

for (int n = 1; n <= 100; n++)
    WriteLine ("{0,3}: {1}", n, FizzBuzz.test(n));

WriteLine ($"\r\nOK @ {DateTime.Now}");

static class FizzBuzz
{
    public enum Category: uint { NONE = 0, FIZZ = 0b0001, BUZZ = 0b0010, FITZBUZZ = 0b0011 };

    private static Category fizz (int n) => n % 3 == 0 ? Category.FIZZ : Category.NONE;
    private static Category buzz (int n) => n % 5 == 0 ? Category.BUZZ : Category.NONE;

    public  static Category test (int n) => fizz (n) | buzz (n);
}

Answer 45

The amount of people here using the most complex and over the top ways to do this very simple task hurts my head. This problem does not need three if statements, something the OP did get right. This was my quick solution:

    for (int i = 1; i<=100; ++i) {
        string printValue = string.Empty;
        printValue += (i%3==0)? "Fizz" : string.Empty;
        printValue += (i%5==0)? "Buzz" : string.Empty;
        if (printValue != string.Empty) Console.WriteLine(printValue);
    }

It avoids using nested loops, there are only two situations to check and storing them in a tuple array is adding complexity to the code with little if any performance or readability gain, but otherwise you had a unique take on it. Im interested if you have programmed in another language, since you used ++i over i++ in your for loop, both work the same in a for loop but the pre increment is slightly more performant, I didn't learn this for a very long time.

As for my code, you can technically just put the condition check into an if statnemt to avoid adding string.Empty to print value if you so wish, but I like the look of this format. You can also more than likely do it in only 3 or so lines, but it gets hard to read.

Now for anyone who has or wants to provide a solution of their own, you can easily simplify this down and use a lookup table or something like that, but the task that was asked was to check EACH number and print a value based on its divisibility. As a programmer, its important to focus on what was asked, not what you think is better. Too many times we will get caught up in trying to outdo ourselves or each other and end up producing something that is either-

• A- Unreadable
• B- Unextendible
• C- Incorrect
• D- Not what the person actually asked for

Answer 46

As of C# 8.0 you can use tuple pattern matching to solve FizzBuzz.

foreach (var item in Enumerable.Range(1, 100).Select(FizzBuzz))
{
    Console.WriteLine(item);
}

static string FizzBuzz(int number)
{
    return (number % 3, number % 5) switch
    {
        (0, 0) => "FizzBuzz",
        (0, _) => "Fizz",
        (_, 0) => "Buzz",
        _ => number.ToString(),
    };
}

Sample based upon this answer.

Answer 47

The shortest answer!!!

have a read.... look i mean:

for(;++$i<101;)print($i%15?$i%3?$i%5?$i:Buzz:Fizz:FizzBuzz)."\n"

Language PHP. Enjoy.