Day 7 of 30 Ruby Coding Challenge: Factorial Numbers

Hey folks!

In today’s post, which is the written version of the Youtube video, we’ll rewrite the previous algorithm with a little bit better version. It’ll be really quick, I promise.

#1 — Algorithm Version 1...As Usual, Ugly

Just to remember, this was the first algorithm version.

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def factorial(number)

  result = number

  while number > 1

    result = result * (number — 1)

    number = number — 1

  end

  return result

end

puts factorial(1)

#2 — Algorithm Version 2. A Little Better

I’d like to keep the argument number intact, which means that I don’t want to use it to control my while loop.

Ruby has an amazing and easy way to deal with arrays in this case:

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(1..(number — 1))

With the code above, we’re saying to Ruby to iterate with numbers between 1 and number — 1.

Notice that -1 is the pattern used for each iteration.

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(1..(number — 1)).each do |item|

end

Notice that the method, each, gives us the current number, which is the item, for each iteration

The final version would be:

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def factorial_version_2(number)

  final_result = number

  (1..(number — 1)).each do |item|

    final_result = final_result * item

  end

  return final_result

end

puts factorial_version_2(5)

As you can see, we avoid the number mutation, which means that our code is less error-prone and easier to read. Cool!

In the next video, we’ll refactor this code a little bit more to reflect a more Ruby way to solve things. See you there :)

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