sublist.exs

defmodule Sublist do
  @doc """
  Returns whether the first list is a sublist or a superlist of the second list
  and if not whether it is equal or unequal to the second list.
  """
  def compare(a, b) do
    cond do
      a == b -> :equal
      a == [] -> :sublist
      b == [] -> :superlist
      sublist_of(a, b) -> :sublist
      sublist_of(b, a) -> :superlist
      true -> :unequal
    end
  end

  # Entry point
  defp sublist_of(a, b), do: sublist_of(a, b, a, false)

  # Start recursing, working through lists together
  # If first item of each list is equal, assume we're starting a match
  # Change 3rd param (found match boolean) and continue on
  defp sublist_of([head_a | tail_a], [head_b | tail_b], a, false) do
    case head_a === head_b do
      true -> sublist_of(tail_a, tail_b, a, true)
      false -> sublist_of([head_a | tail_a], tail_b, a, false)
    end
  end

  # Match has been started, keep checking to see if it continues
  defp sublist_of([head_a | tail_a], [head_b | tail_b], [head_oa | tail_oa], true) do
    cond do
      head_a === head_b -> sublist_of(tail_a, tail_b, [head_oa | tail_oa], true)
      head_oa === head_b -> sublist_of(tail_oa, tail_b, [head_oa | tail_oa], true)
      true -> sublist_of([head_oa | tail_oa], tail_b, [head_oa | tail_oa], false)
    end
  end

  # If we made it this far, then list A was sublist of list B
  defp sublist_of([], [_ | _], _, true), do: true
  defp sublist_of([], [], _, true), do: true
  defp sublist_of([], _, _, false), do: false
  defp sublist_of(_, [], _, _), do: false
end