Find all occurrences of 1 or 2 letters in a string using ruby

Question

If I have a string such as 'abcde' and I want to get a 2d array of all combinations of 1 or 2 letters.

[ ['a', 'b', 'c', 'd', 'e'], ['ab', 'c', 'de'], ['a', 'bc', 'd', 'e'] ...

How would I go abouts doing so?

I want to do this in ruby, and think I should be using a regex. I've tried using

strn = 'abcde'
strn.scan(/[a-z][a-z]/)

but this is only going to give me the distinct sets of 2 characters

['ab', 'cd']

Answer 1

I think this should do it (haven't tested yet):

def find_letter_combinations(str)
  return [[]] if str.empty?
  combinations = []
  find_letter_combinations(str[1..-1]).each do |c|
    combinations << c.unshift(str[0])
  end
  return combinations if str.length == 1
  find_letter_combinations(str[2..-1]).each do |c|
    combinations << c.unshift(str[0..1])
  end
  combinations
end

Answer 2

Regular expressions will not help for this sort of problem. I suggest using the handy Array#combination(n) function in Ruby 1.9:

def each_letter_and_pair(s)
  letters = s.split('')
  letters.combination(1).to_a + letters.combination(2).to_a
end

ss = each_letter_and_pair('abcde')
ss # => [["a"], ["b"], ["c"], ["d"], ["e"], ["a", "b"], ["a", "c"], ["a", "d"], ["a", "e"], ["b", "c"], ["b", "d"], ["b", "e"], ["c", "d"], ["c", "e"], ["d", "e"]]

Answer 3

No, regex is not suitable here. Sure you can match either one or two chars like this:

strn.scan(/[a-z][a-z]?/)
# matches: ['ab', 'cd', 'e']

but you can't use regex to generate a (2d) list of all combinations.

Answer 4

A functional recursive approach:

def get_combinations(xs, lengths)
  return [[]] if xs.empty?
  lengths.take(xs.size).flat_map do |n|
    get_combinations(xs.drop(n), lengths).map { |ys| [xs.take(n).join] + ys } 
  end
end

get_combinations("abcde".chars.to_a, [1, 2])
#=> [["a", "b", "c", "d", "e"], ["a", "b", "c", "de"], 
#    ["a", "b", "cd", "e"], ["a", "bc", "d", "e"], 
#    ["a", "bc", "de"], ["ab", "c", "d", "e"], 
#    ["ab", "c", "de"], ["ab", "cd", "e"]]