So I managed to get part 1 of the day, but it took 2 seconds to run on the real input, which is a bad sign.

I can’t see any kind of optimisation that means I can skip checks and know how many combinations are in those skipped checks (aside from 0.) I can bail out of branches of combinations if the info so far won’t fit, but that still leads me to visiting every valid combination which in one of the examples is 500k. (And probably way more in the input, since if I can’t complete the example near instantly the input is not happening.)

Right now I take the string, then replace the first instance of a ? with the two possible options. Check it matches the check digits so far then use recursion on those two strings.

I can try to optimise the matching, but I don’t think that solves the real problem of visiting every combination.

I don’t think (or hope) it’s just bad code but this is my code so far (python.)

edit:

spoiler

a cache was the solution!

  • Sekoia
    3·
    1 year ago

    I went with a completely different approach:

    ! iterate over our string. Whenever you hit a non-empty, check if the next N are also possible to be a # (N being the first element of our sequence) and that the N+1th isn’t a #. If they are, we can truncate the first N+1, the first element of our sequence, and recurse. If you hit a #, you know that the first element has to start here at the latest, so you can break. With this method, memoization is enough to get part 2 down to 25 ms. To make the memoization more efficient you can also truncate all the way up to the next non-empty when recursing. !<