Uiua

A Christmas Day treat: a one-liner for you all to decipher!

"#####\n.####\n.####\n.####\n.#.#.\n.#...\n.....\n\n#####\n##.##\n.#.##\n...##\n...#.\n...#.\n.....\n\n.....\n#....\n#....\n#...#\n#.#.#\n#.###\n#####\n\n.....\n.....\n#.#..\n###..\n###.#\n###.#\n#####\n\n.....\n.....\n.....\n#....\n#.#..\n#.#.#\n#####"
/+♭⊞(/×<8+)∩°□°⊟ ⊕(□≡≡/+⌕@#)≠@#≡(⊢⊢). ⊜(⍉⊜∘⊸≠@\n)¬±⦷"\n\n". 

Dart

Quick and dirty, and slightly tipsy, code.

Happy Christmas everyone!

Thanks to Eric and the team at Advent of Code, to @Ategon@programming.dev and @CameronDev@programming.dev for giving us somewhere to share and discuss our solutions, and to everyone here for the friendly and supportive community.

See you all next year!

import 'package:collection/collection.dart';
import 'package:more/more.dart';

part1(List<String> lines) {
  var (w, h) = (lines.first.length, lines.indexOf(''));
  var (falsey: keys, truthy: locks) = (lines..insert(0, ''))
      .splitBefore((l) => l.isEmpty)
      .map((g) => [
            for (var x in 0.to(w)) [for (var y in 1.to(h + 1)) g[y][x]]
          ])
      .partition((g) => g[0][0] == '#');
  return keys
      .map((l) => locks.count((k) =>
          0.to(w).every((r) => (l[r] + k[r]).count((e) => e == '#') < 8)))
      .sum;
}

Kotlin

A fun and small challenge. First read all locks, transpose their profile and count the #s (-1 for the full row). Then do the same for the keys.

Lastly find all keys for all locks that do not sum to more than 5 with their teeth:

val lockRegex = Regex("""#{5}(\r?\n[.#]{5}){6}""")
val keyRegex = Regex("""([.#]{5}\r?\n){6}#{5}""")

fun parseLocksAndKeys(inputFile: String): Pair<List<IntArray>, List<IntArray>> {
    val input = readResource(inputFile)
    val locks = lockRegex
        .findAll(input)
        .map {
            it
                .value
                .lines()
                .map { line -> line.toList() }
                .transpose()
                .map { line -> line.count { c -> c == '#' } - 1 }
                .toIntArray()
        }
        .toList()

    val keys = keyRegex
        .findAll(input)
        .map {
            it
                .value
                .lines()
                .map { line -> line.toList() }
                .transpose()
                .map { line -> line.count { c -> c == '#' } - 1 }
                .toIntArray()
        }
        .toList()

    return locks to keys
}

fun part1(inputFile: String): String {
    val (locks, keys) = parseLocksAndKeys(inputFile)

    val matches = locks.map { lock ->
        keys.filter { key ->
            for (i in lock.indices) {
                // Make sure the length of the key and lock do not exceed 5
                if (lock[i] + key[i] > 5) {
                    return@filter false
                }
            }
            true
        }
    }
        .flatten()
        .count()

    return matches.toString()
}

Also on GitHub

Haskell

Merry Christmas!

{-# LANGUAGE OverloadedStrings #-}

module Main where

import Data.Either
import Data.Text hiding (all, head, zipWith)
import Data.Text qualified as T
import Data.Text.IO as TIO

type Pins = [Int]

toKeyLock :: [Text] -> Either Pins Pins
toKeyLock v = (if T.head (head v) == '#' then Left else Right) $ fmap (pred . count "#") v

solve keys locks = sum [1 | k <- keys, l <- locks, fit k l]
  where
    fit a b = all (<= 5) $ zipWith (+) a b

main = TIO.getContents >>= print . uncurry solve . partitionEithers . fmap (toKeyLock . transpose . T.lines) . splitOn "\n\n"

C

Merry Christmas everyone!

#include "common.h"

int
main(int argc, char **argv)
{
	static char buf[7];
	static short h[500][5];	/* heights */
	static short iskey[500];
	int p1=0, nh=0, i,j,k;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	
	for (nh=0; !feof(stdin) && !ferror(stdin); nh++) {
		assert(nh < (int)LEN(h));

		for (i=0; i<7; i++) {
			fgets(buf, 7, stdin);
			if (i==0)
				iskey[nh] = buf[0] == '#';
			for (j=0; j<5; j++)
				h[nh][j] += buf[j] == '#';
		}

		/* skip empty line */
		fgets(buf, 7, stdin);
	}

	for (i=0; i<nh; i++)
	for (j=0; j<nh; j++)
		if (iskey[i] && !iskey[j]) {
			for (k=0; k<5 && h[i][k] + h[j][k] <= 7; k++) ;
			p1 += k == 5;
		}

	printf("25: %d\n", p1);
	return 0;
}

https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day25.c

Made the 1 second challenge with most of it to spare! 😎

$ time bmake bench                                                                                                      
day01  0:00.00  1912 Kb  0+88 faults                                                                                            
day02  0:00.00  1992 Kb  0+91 faults 
day03  0:00.00  1920 Kb  0+93 faults
day04  0:00.00  1912 Kb  0+90 faults 
day05  0:00.00  2156 Kb  0+91 faults
day06  0:00.03  1972 Kb  0+100 faults
day07  0:00.06  1892 Kb  0+89 faults
day08  0:00.00  1772 Kb  0+87 faults 
day09  0:00.02  2024 Kb  0+137 faults
day10  0:00.00  1876 Kb  0+87 faults 
day11  0:00.00  6924 Kb  0+3412 faults
day12  0:00.00  1952 Kb  0+103 faults
day13  0:00.00  1908 Kb  0+88 faults
day14  0:00.05  1944 Kb  0+92 faults                                                                                            
day15  0:00.00  2040 Kb  0+89 faults
day16  0:00.03  2020 Kb  0+250 faults
day17  0:00.00  1896 Kb  0+88 faults
day18  0:00.00  1952 Kb  0+107 faults
day19  0:00.01  1904 Kb  0+91 faults
day20  0:00.01  2672 Kb  0+325 faults
day21  0:00.00  1804 Kb  0+86 faults
day22  0:00.03  2528 Kb  0+371 faults
day23  0:00.02  2064 Kb  0+152 faults
day24  0:00.00  1844 Kb  0+89 faults
day25  0:00.00  1788 Kb  0+89 faults  
                                                                
real    0m0,359s