Solving Fizzbuzz Using Rubys Enumerable Module?

Question

Hey I was given the fizzbuzz task recently and I had answered with the usual,

if ((i%3==0) || (i.to_s.include?('3'))) && ((i%7==0) || (i.to_s.include?('7')))
p 'Fizzbuzz'
elsif (i%3==0) || (i.to_s.include?('3'))
p 'Fizz'
elsif (i%7==0) || (i.to_s.include?('7'))
p 'Buzz'
else
p i
end

and when asked to shorten it I tried using ternary operators:

p (i%3<1 || i.to_s.include?('3')) ? ((i%7<1 || i.to_s.include?('7')) ? "Fizzbuzz" : "Fizz") : ((i%7<1 || i.to_s.include?('7')) ? "Buzz" : i)

but when asked to solve it using the Enumerable methods(select,reject,collect etc) I was well stumped...Any body tried this before??

The select/collect methods were specificaly mentioned so I'm guessing that he had something like this in mind(excuse the crappy code) (1..100).select { |i| i % 3 == 0 }.collect { "fizz" } but i'm stuck when trying to do this for the 3 conditions and print out the result(ie iterate through the output array) :\

It is generally considered a bad practice to nest ternary operator expressions. — sawa, May 28 '13 at 23:38
What are you doing with expressions like `i.to_s.include?('3')`? Do you want numbers like `31` to be classified together with multiples of 3? — sawa, May 28 '13 at 23:41
The usual? Just creating a string for every number to compare it with another number converted to a string? Well, in my time ... — steenslag, May 28 '13 at 23:46
Yes, the solution needs to return Fizz if the number contains or is a multiple of 3 etc.. — Conor, May 29 '13 at 00:07
Are you sure he didn't mean that he wanted you to include the code for enumerating over the numbers rather than use Enumerable for deciding what to print? — Chuck, May 29 '13 at 00:14
Off topic, but still fun to [look](http://codegolf.stackexchange.com/questions/88/obfuscated-fizzbuzz-golf) at. — squiguy, May 29 '13 at 00:21

score 2 · Answer 1 · answered May 29 '13 at 01:40

Probably not much help now, but I've produced a gem (fizzbuzzard) that monkey patches Fixnum such that all multiples of three print as Fizz, all multiples of five etc etc. Interviewers will respect you for using your knowledge of existing libraries rather than pointlessly re-solving solved problems.

Get it from rubgems here and find the source here.

The part I'm proudest of? The test suite. Output reproduced here:

(0h8m|master) [fizzbuzzard] $ rspec
.

Finished in 0 seconds
1 example, FIZZBUZZ failures

Shawn Balestracci · Answer 2 · 2013-05-29T00:24:03.263

1

def fizzbuzz(i)
  [nil,nil,"Fizz",nil, nil,nil,"Buzz"].each_with_index do |fizz_or_buzz, idx|
    print fizz_or_buzz if (i.to_i % (1+idx)).zero?
  end
end

If the interviewer doesn't laugh, then it's probably not going to be a good fit (for me anyways).

Note that print doesn't add a newline (vs p)

See http://rosettacode.org/wiki/FizzBuzz#Ruby for more common solutions.

edited May 29 '13 at 00:24

answered May 28 '13 at 23:49

Shawn Balestracci

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`NameError: undefined local variable or method 'i' for main:Object` – Daniel X Moore May 29 '13 at 00:00
1

I asssumed `i` was set beforehand (as in the original question) @DanielXMoore – Shawn Balestracci May 29 '13 at 00:01

score 1 · Answer 3 · answered May 29 '13 at 00:26

This one's a little dense but fun. It abuses arrays and some functional methods slightly to avoid using any if statements.

rules = {
  3 => ["fizz"],
  7 => ["buzz"]
}

(1..100).map do |n|
  [rules.map do |key, value|
    value[n % key]
  end.compact.join, n.to_s].find do |word|
    !word.empty?
  end
end.join("\n")

score 0 · Answer 4 · answered May 29 '13 at 01:30

It is not clean to repeat the same conditionals.

p case [
  i.%(3).zero? || i.to_s.include?("3"),
  i.%(7).zero? || i.to_s.include?("7")
]
when [true, true] then "Fizzbuzz"
when [true, false] then "Fizz"
when [false, true] then "Buzz"
when [false, false] then i
end

pjs · Answer 5 · 2013-05-29T13:46:06.397

0

This hands you back the results in an array of strings, you can do whatever you want with them at that point.

results = (1..100).to_a.collect do |i|
  f = ((i%3).zero? || i.to_s.include?('3'))
  b = ((i%7).zero? || i.to_s.include?('7'))
  "#{:fizz if f}#{:buzz if b}#{i unless f || b}"
end
p results

edited May 29 '13 at 13:46

answered May 29 '13 at 01:51

pjs

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Solving Fizzbuzz Using Rubys Enumerable Module?

5 Answers5