This is what I ended up with, which I think is relatively clear and
straightforward (but then, I'm not entirely unbiased :)
The algorithm is very close to what you described.
- Generate all sub-sequences of length > 1
- Filter to keep only increasing subsequences
- Tack on the empty sequence, which is what needs to be returned if there
are no increasing sub-sequences
- Sort by length in descending order
- Take the first (longest)
(fn [coll]
(let [increasing? (fn [xs] (apply < xs))
n (count coll)
sub-seqs (mapcat #(partition % 1 coll) (range 2 (inc n)))]
(->> sub-seqs
(filter increasing?)
(cons [])
(sort-by count >)
first)))
On Tuesday, June 12, 2012 12:11:07 PM UTC-7, Andy C wrote:
>
> Hi,
>
> First a quick disclaimer. Those are my first steps in Clojure so I am
> not be super accustomed to the language entire landscape and might
> miss some basics here. However I was able to solve my first 4clojure
> hard problem https://www.4clojure.com/problem/53 and have some second
> thoughts after looking up top contributor's solutions as well as mine.
> Why it has to be so complicated???
>
> Conceptually, you just reduce the list to list of lists using a simple
> condition < . Then you filter items longer then 1. And at the same
> time you reduce the output to a first longest list. In this case,
> stack for recursion is really not required, although I did use it in
> my solution since I could figure out the "reduction" based way to
> partition the source sequence.
>
> It also seems that imperative solution would be quite straightforward
> although maintaining at least 4 state variables is not compelling at
> all. Bottom line, I want to have a idiomatic Clojure solution ... Any
> insight ....
>
> Thx,
> Andy
>
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