Kevin Compton wrote:
> I think a big part of it is that there are variations in what we assume 
> the
> questioner wanted.
> 
> In my case, I interpreted it as:
>  1) remove items that appear in both lists from both lists
>      For instance,  removeDups ([a, b, c, d], [b, d, f, g]) => [a, c], 
> [f,
> g]
>  2) don't go so far as remove more than the common count of dups
>      For instance, removeDups ([a, a, b, b, c, d, d], [b, d, d, d, f, g, 
> g])
> => [a, a, b, c], [d, f, g, g]
>  3) keep the lists in original order (probably not required but I'm not
> sure)
> 
> I guess, it would be nice to have had the *need* demonstrated via 
> example or
> clearly stated..
> but its been fun.

I have to say that I am almost certainly being simplistic here but why 
cannot we do something like this:

a = [1, 2, 3, 4]
b = [2, 4, 6, 8]

p a
p b

c = a & b
a = a - c
b = b - c

p c
p a
p b

result:
[1, 2, 3, 4]
[2, 4, 6, 8]
[2, 4]
[1, 3]
[6, 8]

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