ABC Problem

# title: ABC Problem
# source: https://rosettacode.org/wiki/ABC_Problem
# category: Rosetta Code Popular
# difficulty: Intermediate
# summary: Determine whether a word can be formed from a fixed set of two-letter blocks without reusing a block.
# tags: popular, search, strings, recursion
# vibe: 0.4

def blocks
  [
    "BO", "XK", "DQ", "CP", "NA",
    "GT", "RE", "TG", "QD", "FS",
    "JW", "HU", "VI", "AN", "OB",
    "ER", "FS", "LY", "PC", "ZM"
  ]
end

def block_matches?(block, letter)
  block.start_with?(letter) || block.slice(1) == letter
end

def can_make_from?(letters, letter_index, available_blocks)
  if letter_index >= letters.size
    return true
  end

  letter = letters[letter_index]
  block_index = 0

  while block_index < available_blocks.size
    if block_matches?(available_blocks[block_index], letter)
      remaining = available_blocks.take(block_index) + available_blocks.drop(block_index + 1)
      if can_make_from?(letters, letter_index + 1, remaining)
        return true
      end
    end
    block_index = block_index + 1
  end

  false
end

def can_make?(word)
  can_make_from?(word.upcase.chars, 0, blocks)
end

def run
  {
    a: can_make?("A"),
    bark: can_make?("BARK"),
    book: can_make?("BOOK"),
    treat: can_make?("TREAT"),
    common: can_make?("COMMON"),
    squad: can_make?("SQUAD"),
    confuse: can_make?("CONFUSE")
  }
end