David Jagoe <davidja...@gmail.com> writes:
Hi David,
> Thanks! I had a nasty feeling that it could be done in a one-liner
> using partition-by or something similar. But of course you need to be
> able to look at previous elements...
Here's a slightly shorter variant.
--8<---------------cut here---------------start------------->8---
(defn partition-distinct [c]
(reduce #(if (.contains (last %1) %2)
(conj %1 [%2])
(conj (subvec %1 0 (dec (count %1)))
(conj (last %1) %2)))
[[]] (vec c)))
--8<---------------cut here---------------end--------------->8---
The difference is that it'll always return a vector of vectors, and of
course the .contains-check is linear for vectors, so it's quadratic to
the size of `c` in total.
If you don't care about the order of elements in each partition, then
you can just go with a vector of sets and have the same performance as
Cedric's recursive solution.
--8<---------------cut here---------------start------------->8---
(defn partition-distinct [c]
(reduce #(if ((last %1) %2)
(conj %1 #{%2})
(conj (subvec %1 0 (dec (count %1)))
(conj (last %1) %2)))
[#{}] (vec c)))
--8<---------------cut here---------------end--------------->8---
Bye,
Tassilo
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