adding array elements using recursion

Question

I need to create a recursive function that adds the numbers of any given array, then removes the first element, then adds the array and do this until the array only has element left. my function does this but at the moment I can only see the addition by putting a puts statement but also need the results as the return value like this =>[20,20,19,16,10] and dont know how to go about this as its putting the results separately. Thanks for your help. The function needs to do this recursively: I this is my code:

def parts_sums(ls)
    results_arr=[]
    if( ls.length === 1)
     return ls 
    end
     p results =ls.sum
     parts_sums(ls.drop(1))
     p results_arr << results

end
parts_sums([0, 1, 3, 6, 10])

# ls = [0, 1, 3, 6, 10].sum =>20
# ls = [1, 3, 6, 10].sum => 20
# ls = [3, 6, 10].sum =>19
# ls = [6, 10].sum =>16
# ls = [10]=>10

puts returns nil. You're first and foremost concern when writing any method should be its return value. You also don't really need a recursive method unless you're writing it to learn about recursiveness as you can do this with each_with_object. — max
– max, Commented Apr 22, 2020 at 11:36
I removed the ruby-on-rails tag. This is purely a Ruby question and doesn't pertain to the Ruby on Rails framework. — lurker
– lurker, Commented Apr 22, 2020 at 12:30

iGian · Accepted Answer · 2020-04-22 13:22:29Z

2

You can define a method like this, which is verbose but clear (I think):

def parts_sums(ary, res = [])
  res << [ary.dup, ary.sum]
  if ary.size > 1
    ary.shift
    parts_sums(ary, res)
  else
    return res
  end
end

So, when you call on your array, you get this result:

ary = [0, 1, 3, 6, 10]
parts_sums(ary)
#=> [[[0, 1, 3, 6, 10], 20], [[1, 3, 6, 10], 20], [[3, 6, 10], 19], [[6, 10], 16], [[10], 10]]

Call parts_sums(ary.dup) if you want to preserve the original array.

Which can be rewritten in this shortest way:

def parts_sums_2(ary, res = [])
  return res unless ary.any?
  res << [ary, ary.sum]
  parts_sums_2(ary[1..-1], res)
end

edited Apr 22, 2020 at 13:22

answered Apr 22, 2020 at 13:05

iGian

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Comments

Pavel Oganesyan · Accepted Answer · 2020-04-22 11:38:27Z

1

def parts_sums(ls)
  ls.length == 1 ? ls : ([ls.sum] + parts_sums(ls[1..-1]))
end

puts parts_sums([0, 1, 3, 6, 10]).to_s

answered Apr 22, 2020 at 11:38

Pavel Oganesyan

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3 Comments

Fabio Andres Over a year ago

Thanks Pavel. This part here (ls[1..-1]) What does it do please?

Pavel Oganesyan Over a year ago

@FabioAndres It takes elements from ls array starting with index 1 (so, element with index 0 will not be included in result) and to the end: -1 here marks the last element. So, it just takes all elements except the first one. ls[1..2] would take the second and the third elements.

Stefan Over a year ago

Calling Array#sum in a recursive method doesn't seem right.

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