flatten2(List) -> reverse( flatten_helper(List,[]) ).
flatten_helper([],Dst) -> Dst;
flatten_helper([H|T],Dst) when is_list(H) -> flatten_helper( concatenate([H,T]), Dst );
flatten_helper([H|T],Dst) -> flatten_helper( T, [H|Dst] ).

you could avoid is_list by matching [[HH|HT]|T] and [[]|T] separately as in your example.

I am now trying to break it

Found your blog while doing the exercises and searching for the answer to a question I had about 3.5 (writing flatten). Basically it says to use “concatenate to solve flatten” and even though I solved flatten similarly to the first way you did (my code below), I also didn’t use concatenate.

What I’m OCDing on is I actually think his hint on using concatenate is wrong here, because erlang’s lists are untyped (so you never know if your dealing with a list of lists (which is what concatenate takes) until you navigate it).

Regardless, great blog, and I’m going to start following your solutions to his book.

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flatten([], Accum) ->
Accum;
flatten([H|[]], Accum) ->
flatten(H, Accum);
flatten([H|T], Accum) ->
flatten(H, flatten(T, Accum));
flatten(Val, Accum) ->
[Val | Accum].

flatten(List) ->
flatten(List,[]).