advent-of-code VS AdventOfCode2021

Rust GitHub. My approach is to go through all cuboids and always remove their intersection with all previously added cuboids. Then, If it is a "on" cuboid it is added to the list of cuboids. This way no cuboids will overlap and I can sum all their volumes in the end.

Rust GitHub. Kinda ugly brute force solution which runs in ~10 sec.

Rust GitHub. This took me much longer that I care to admit. I started with a tree-based solution, tried a list, went back to tree and the back to the list again...

Rust GitHub

GitHub.

Rust GitHub. This took a very long time to solve, and the code is not very nice looking.

Full program on GitHub.

Full program including code to deal with heaps, on GitHub.

Program I used for the above results

See my solution on GitHub.

Note also that you can solve this without a complexity dependency on n. Most solutions I have seen, including mine run in time O(tD), where t is the maximum value of a timer, and D the number of generation, requiring O(t) memory. (This is assuming we can do the required arithmetic operations in constant time, and each numbers require a fixed around of memory storage; if the number of fish gets huge, throw in an additional log n in the complexities, where n is the number of fish on the final day).

Based on my matrix exponentiation solution, here is a closed-form solution:

In my Perl solution of today (I assume, you mean 2021, Day 4), I read stuff in paragraph mode ($/ = "") which makes Perl split input on 2 or more newlines.

I considered doing that for my (Perl) solution, but given the cards are small, that seemed overkill.

Python implementation on GitHub