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@Kerrick
Created April 24, 2012 20:36
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Different solutions for Fizz Buzz in Ruby
def fizz_buzz_1(max)
arr = []
(1..max).each do |n|
if ((n % 3 == 0) && (n % 5 == 0))
arr << "FizzBuzz"
elsif (n % 3 == 0)
arr << "Fizz"
elsif (n % 5 == 0)
arr << "Buzz"
else
arr << n
end
end
return arr
end
def fizz_buzz_2(max)
arr = []
(1..max).each do |n|
if (n % 3 == 0)
if (n % 5 == 0)
arr << "FizzBuzz"
else
arr << "Fizz"
end
elsif (n % 5 == 0)
arr << "Buzz"
else
arr << n
end
end
return arr
end
def fizz_buzz_3(max)
arr = []
(1..max).each do |n|
text = ""
if (n % 3 == 0)
text << "Fizz"
end
if (n % 5 == 0)
text << "Buzz"
end
if !((n % 3 == 0) || (n % 5 == 0))
text = n
end
arr << text
end
return arr
end
@ragesoss
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ragesoss commented Jul 7, 2016

class Fixnum
  def to_s
    string = ''
    string += 'fizz' if self % 3 == 0
    string += 'buzz' if self % 5 == 0
    return string unless string.empty?
    inspect
  end
end

puts (1..100).to_a

@algex
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algex commented Oct 20, 2016

(1..100).each do |i|
if i % 5 == 0 && i % 3 == 0
puts 'FizzBuzz'
elsif i % 5 == 0
puts 'Buzz'
elsif i % 3 == 0
puts 'Fizz'
else
puts i
end

end

@g-mehra
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g-mehra commented Jan 16, 2017

#FizzBuzz

(1..100).each do |number|
if (number % 3 == 0) and (number % 5 != 0)
puts "Fizz"
elsif (number % 5 == 0 ) and (number % 3 != 0)
puts "Buzz"
elsif (number % 5 == 0) and (number % 3 == 0)
puts "FizzBuzz"
else
puts number
end
end

@besya
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besya commented Mar 9, 2017

buzzer = lambda do |i|
  map = {ff: i, tf: 'Fizz', ft: 'Buzz', tt: 'FizzBuzz'}
  map[((i%3==0).to_s[0] + (i%5==0).to_s[0]).to_sym]
end
puts (1..100).map(&buzzer)

@juhaniahola
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juhaniahola commented May 5, 2017

(0..100).map{|i|i%15==0?'FizzBuzz':i%3==0?'Fizz':i%5==0?'Buzz':i}

#66 chars :)

@dankohn
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dankohn commented Dec 31, 2017

This one requires presence from Rails (i.e., it works in rails console):

puts 1.upto(100).map { |n| "#{'Fizz' if n%3==0}#{'Buzz' if n%5==0}".presence || n }

@Aketzu
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Aketzu commented Feb 13, 2018

(1..100).map{|n|({3=>['Fizz'],5=>['Buzz']}.map{|x,y|y[n%x]}.join.gsub(/^$/,n.to_s))}

@kke
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kke commented May 3, 2018

@willwright82 's solution modified for less == 0:

(1..100).map do |m|
  case 0
  when m % 15 then 'FizzBuzz'
  when m % 3  then 'Fizz'
  when m % 5  then 'Buzz'
  else m
  end
end

@coryjanowski
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def fizzbuzz(int)
if int % 3 == 0 && int % 5 == 0
"FizzBuzz"
elsif int % 3 == 0
"Fizz"
elsif int % 5 == 0
"Buzz"
else int % 4 == 0
nil
end
end

@lux9
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lux9 commented Oct 5, 2020

70 characters

(1..100).map{|x|puts x%15==0?'FizzBuzz':x%5==0?'Buzz':x%3==0?'Fizz':x}

@khamusa
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khamusa commented Apr 29, 2021

When I thought we already had way too many solutions... !

module BeingMultiple
  refine Integer do
    def eitherFizzOrBuzz!
      return "FizzBuzz" if multiple?(3) && multiple?(5)
      return "Fizz" if multiple?(3)
      return "Buzz" if multiple?(5)
      self
    end
    
    private
    
    def multiple?(target)
      self % target == 0
    end
  end
end

using BeingMultiple

puts (1..n).map(&:eitherFizzOrBuzz!).join("\n")

@al3rez
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al3rez commented Sep 7, 2021

require "fizzbuzz"

RSpec.describe FizzBuzz do
  describe "#compute" do
    context "when number is dvisisble by 3" do
      it "returns fizz" do
        expect(FizzBuzz.compute(6)).to eq("fizz")
      end
    end

    context "when number is dvisisble by 5" do
      it "returns buzz" do
        expect(FizzBuzz.compute(10)).to eq("buzz")
      end
    end

    context "when number is dvisisble by both 3 and 5" do
      it "returns buzz" do
        expect(FizzBuzz.compute(15)).to eq("fizzbuzz")
      end
    end
  end
end


class FizzBuzz
  def self.compute(number)
    return "fizzbuzz" if (number % 15).zero?

    if (number % 3).zero? "fizz" else "buzz" end
  end
end

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