Heres my go. Originally I had something that was performing between 2N-1 and N*((N+1)/2), but kaif provided the insight I needed to get it to N log N when he pointed out that you only need to store one path of a given length at any one time.

BTW, I wanted a single routine that could return either the indexes or the actual values, so this is a little clunkier than it need be if you didnt care about the indexes.

Update: I was wrong, the worse case run time for this routine is (N+1)/2*N, which is O(N2).

<Reveal this spoiler or all in this thread>
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$world=~s/war/peace/g

In reply to Re: Puzzle: Longest Increasing Sequence by demerphq
in thread Puzzle: Longest Increasing Sequence by TedPride

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