Given a set of strings, calculate the set of unambiguous abbreviations for those strings, and return a hash where the keys are all the possible abbreviations and the values are the full strings. Thus, given input of "car" and "cone", the keys pointing to "car" would be "ca" and "car", while those pointing to "cone" would be "co", "con", and "cone".

The optional pattern parameter is a pattern or a string. Only those input strings matching the pattern, or begging the string, are considered for inclusion in the output hash

Source Code

# File abbrev.rb, line 44
def abbrev(words, pattern = nil)
  table = {}
  seen = Hash.new(0)

  if pattern.is_a?(String)
    pattern = /^#{Regexp.quote(pattern)}/     # regard as a prefix
  end

  words.each do |word|
    next if (abbrev = word).empty?
    while (len = abbrev.rindex(/[\w\W]\z/)) > 0
      abbrev = word[0,len]

      next if pattern && pattern !~ abbrev

      case seen[abbrev] += 1
      when 1
        table[abbrev] = word
      when 2
        table.delete(abbrev)
      else
        break
      end
    end
  end

  words.each do |word|
    next if pattern && pattern !~ word

    table[word] = word
  end

  table
end