My solution:
(defn lis [s]
(->> s
(partition 2 1)
(partition-by (partial apply <=))
(filter (fn [[[a b]]] (< a b)))
(reduce (fn [m s] (if (> (count s) (count m)) s m)) [])
(#(cons (ffirst %) (map second %)))))
The strategy is (given the vector [3 2 1 2 4 2]
- I get the sequence of consecutive pairs in the original vector
((3 2)(2 1)(1 2)(2 4)(4 2))
- I partition in sequences of pairs while the ordering criteria in a pair
is maintained
(((3 2) (2 1))((1 2)(2 4))((4 2)))
- I filter out those with correspond to descendant subsequences
(((1 2)(2 4)))
- I get the maximun lenght sequene. I use reduce to compute it
incrementally
(((1 2)(2 4)))
- I reconstruct the sequence adjoint the first of the first pair with the
seconds of each pair
(1 2 4)
Maybe there are some corner simplifications but I think it works
Juan Manuel
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