How to return an array with cumulative values in a recursive method

Question

I am attempting to take two values times and number and return an array of number repeated times times. Here is an example:

replicate(3, 5) # => [5, 5, 5]

Here is my attempt 1:

@array = []

def replicate(times, number)
  return [] if times <= 0
  @array << number 
  replicate(times - 1, number)
  @array
end

When I run each test case once in isolation, I pass everything. However, when I run them all at once, it fails; @array contains all the values for every test case and @array will look like this at the end of the entire test suite:

@array # => [5, 5, 5, 1, 1, 1, 1, 2, 2]

Here is implementation two:

def replicate(times, number)
  return [] if times <= 0
  array = []
  array << number 
  replicate(times - 1, number)
  array
end

This will return only one value because recursion will create a local copy for every run.

How can I return an array that will make the test cases pass? I can't use global or instance variables, or a local copy of an array. Is there something I can use in between?

Try passing array into your recursive function defination.

Shravan40
– Shravan40

2016-06-27 14:26:11 +00:00
Commented Jun 27, 2016 at 14:26 — Shravan40
– Shravan40, Commented Jun 27, 2016 at 14:26

Jordan Running · Accepted Answer · 2016-06-27 14:38:08Z

5

In this case there's no need to keep an accumulator either as an instance variable or an argument. All you need to do is concatenate the result from this call (which will be an array with a single element, i.e. [ number ]) with the result of subsequent calls:

def replicate(times, number)
  return [] if times <= 0
  [ number ] + replicate(times - 1, number)
end

If you wanted to use an accumulator, you could do it this way:

def replicate(times, number, accum=[])
  return accum if times <= 0
  replicate(times - 1, number, accum + [ number ])
end

(Note that you could use accum << number as well, but personally I prefer to treat data structures as immutable when writing recursive functions—because, well, it's a capital-F function and functions shouldn't have side-effects.)

edited Jun 27, 2016 at 14:38

answered Jun 27, 2016 at 14:29

Jordan Running

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2 Comments

Nic Nilov Over a year ago

This is a better answer.

Jordan Running Over a year ago

Thanks, @NicNilov. Your solution was good as well. Sorry someone downvoted it. :(

Nic Nilov · Accepted Answer · 2016-06-27 14:25:27Z

5

Pass your accumulator array to recursive calls:

def replicate(times, number, array=[])
   return [] if times <= 0
   array << number
   replicate(times - 1, number, array)
   array
end

replicate(3, 5)
=> [5, 5, 5]

answered Jun 27, 2016 at 14:25

Nic Nilov

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2 Comments

sawa Over a year ago

This is a pattern frequently seen. Is there a name for this?

Nic Nilov Over a year ago

Accumulative recursion is one way to call it

Pholochtairze · Accepted Answer · 2016-06-27 14:31:52Z

0

Why not simply do :

[number]*times

Example :

$ irb
2.3.0 :001 > [1]*5
 => [1, 1, 1, 1, 1]

answered Jun 27, 2016 at 14:31

Pholochtairze

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