Complex list (double numbering) in LaTeX

Question

I'm working on a text that has a long list of numbered items (some with sub-numbers). Under normal circumstances, there wouldn't be an issue, but someone (not me) wanted to have cross-references to another set of numbered items. If the items in "my" list are capital letters and the items in the other list are lower-case letters, the top level of the list looks like this:

A (a). B (b). C (f). D (h and u). E (z).

There is no double-labeling of the sub-item lists.

Is there a way to add an (optional) argument to \enumerate that will add a number I give it in parentheses AND eliminate the parenthetical argument if I don't give it that argument? Something like this:

\item First item

\item {32} Second item
\begin{enumerate}
\item Sub-item one
\item Sub-item two
\item Sub-item three
\end{enumerate}

\item {4 and 68} Third item

\item Fourth item
\end{enumerate}

That would, in principle, produce:

1. First item

2 (32). Second item

a) Sub-item one

b) Sub-item two

c) Sub-item three

3 (4 and 68). Third item

4. Fourth item

(Note the location of the full stop after each label and the parentheses in the second-level list.)

Answer 1

It's difficult to do it really generally with \item.

\documentclass{article}
\usepackage{enumitem}

\newlist{bienumerate}{enumerate}{1}
\setlist[bienumerate]{%
  label=\arabic*\additionallabel.,
  wide,
}
\NewDocumentCommand{\additionallabel}{}{}
\NewDocumentCommand{\biitem}{o}{%
  \IfNoValueTF{#1}{%
    \renewcommand{\additionallabel}{}%
  }{%
    \renewcommand{\additionallabel}{ (#1)}%
  }%
  \item
}

\begin{document}

\begin{bienumerate}
\biitem First item
\biitem[32] Second item
\biitem[4 and 68] Third item
\biitem Fourth item
\end{bienumerate}

\begin{bienumerate}[label=\Alph*\additionallabel.]
\biitem First item
\biitem[32] Second item
\biitem[4 and 68] Third item
\biitem Fourth item
\end{bienumerate}

\end{document}