Ruby Challenge: Calculating Factorials with Negative Numbers

Welcome to Ruby Tuesday, a weekly post where we explore the world of Ruby programming language. Today's challenge is all about modifying a factorial method in Ruby to handle negative input numbers.

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In this code, the factorial method takes a number n as input and recursively calculates its factorial. The base case is when n is 0, in which case the method returns 1. Otherwise, the method multiplies n by the factorial of n-1.

When we call factorial(5) using the puts method, it calculates the factorial of 5 (i.e., 5 * 4 * 3 * 2 * 1) and outputs the result, which is 120.

def factorial(n)
if n == 0
1
else
n * factorial(n-1)
end
end

puts factorial(5) # outputs 120

Challenge:

Modify the factorial method to handle negative input numbers.

Put your Ruby skills to the test. Go!

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Leave your solutions, thoughts, tricks, and questions in the comments below. And be sure to join us here on CodeNewbie Org next Ruby Tuesday for more exciting challenges and tips to enhance your Ruby skills!.

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Comments

[Hartmut B.]

That's too easy ....

3.2.0 :021 > def factorial(n)
3.2.0 :022 > if n == 0
3.2.0 :023 > 1
3.2.0 :024 > else
3.2.0 :025 > n>0 ? n * factorial(n-1) : n * factorial(n+1)
3.2.0 :026 > end
3.2.0 :027 > end

q&d off cause

[Oliver]

Great answer.

What answer should it give when the input is -5? I assume -120 is correct?

I've modified your answer to be more idiomatic.

class Integer
  def factorial
    return 1 if self.zero?

    return self * (self+1).factorial if self <= 0

    self * (self-1).factorial
  end
end

By defining the method on Integer, I can now call 5.factorial or -5.factorial.

I can see there is room for re-factoring. Would it be clearer without the duplication?

5.factorial
# => 120
-5.factorial
# => -120
0.factorial
# => 1

[Rachel Fazio]

Love this!