Haskell

My naive solution was taking ages until I tried matching from right to left instead :3

In the end the cache required for part two solved the problem more effectively.

import Control.Arrow
import Control.Monad.State
import Data.List
import Data.List.Split
import Data.Map (Map)
import Data.Map qualified as Map

arrangements :: [String] -> String -> Int
arrangements atoms = (`evalState` Map.empty) . go
  where
    go "" = return 1
    go molecule =
      let computed = do
            c <- sum <$> mapM (\atom -> maybe (return 0) go $ stripPrefix atom molecule) atoms
            modify (Map.insert molecule c)
            return c
       in gets (Map.!? molecule) >>= maybe computed return

main = do
  (atoms, molecules) <- (lines >>> (splitOn ", " . head &&& drop 2)) <$> readFile "input19"
  let result = map (arrangements atoms) molecules
  print . length $ filter (> 0) result
  print . sum $ result

C#

public class Day19 : Solver {
  private string[] designs;

  private class Node {
    public Dictionary<char, Node> Children = [];
    public bool Terminal = false;
  }

  private Node root;

  public void Presolve(string input) {
    List<string> lines = [.. input.Trim().Split("\n")];
    designs = lines[2..].ToArray();
    root = new();
    foreach (var pattern in lines[0].Split(", ")) {
      Node cur = root;
      foreach (char ch in pattern) {
        cur.Children.TryAdd(ch, new());
        cur = cur.Children[ch];
      }
      cur.Terminal = true;
    }
  }

  private long CountMatches(Node cur, Node root, string d) {
    if (d.Length == 0) return cur.Terminal ? 1 : 0;
    if (!cur.Children.TryGetValue(d[0], out var child)) return 0;
    return CountMatches(child, root, d[1..]) + (child.Terminal ? CountMatches(root, d[1..]) : 0);
  }

  private readonly Dictionary<string, long> cache = [];
  private long CountMatches(Node root, string d) {
    if (cache.TryGetValue(d, out var cached_match)) return cached_match;
    long match = CountMatches(root, root, d);
    cache[d] = match;
    return match;
  }

  public string SolveFirst() => designs.Where(d => CountMatches(root, d) > 0).Count().ToString();

  public string SolveSecond() => designs.Select(d => CountMatches(root, d)).Sum().ToString();
}

Python

Approach: Recursive memoized backtracking with a Trie

I get to use one of my favorite data structures here, a Trie! It helps us figure out whether a prefix of the design is a valid pattern in linear time.

I use backtracking to choose potential component patterns (using the Trie), kicking off matching the rest of the design down the stack. We can continue matching longer patterns immediately after the recursion stack unwinds.
In addition, I use global memoization to keep track of the feasibility (part 1) or the number of combinations (part 2) for designs and sub-designs. This way, work done for earlier designs can help speed up later ones too.

I ended up combining part 1 and 2 solutions into a single function because part 1 is a simpler variant of part 2 where we count all designs with the number of possible pattern combinations > 0.

import os

here = os.path.dirname(os.path.abspath(__file__))

# read input
def read_data(filename: str):
    global here

    filepath = os.path.join(here, filename)
    with open(filepath, mode="r", encoding="utf8") as f:
        return f.read()

class Trie:
    class TrieNode:
        def __init__(self) -> None:
            self.children = {}  # connections to other TrieNode
            self.end = False  # whether this node indicates an end of a pattern

    def __init__(self) -> None:
        self.root = Trie.TrieNode()

    def add(self, pattern: str):
        node = self.root
        # add the pattern to the trie, one character at a time
        for color in pattern:
            if color not in node.children:
                node.children[color] = Trie.TrieNode()
            node = node.children[color]
        # mark the node as the end of a pattern
        node.end = True

def soln(filename: str):
    data = read_data(filename)
    patterns, design_data = data.split("\n\n")

    # build the Trie
    trie = Trie()
    for pattern in patterns.split(", "):
        trie.add(pattern)

    designs = design_data.splitlines()

    # saves the design / sub-design -> number of component pattern combinations
    memo = {}

    def backtrack(design: str):
        nonlocal trie

        # if design is empty, we have successfully 
        #   matched the caller design / sub-design
        if design == "":
            return 1
        # use memo if available
        if design in memo:
            return memo[design]

        # start matching a new pattern from here
        node = trie.root
        # number of pattern combinations for this design
        pattern_comb_count = 0
        for i in range(len(design)):
            # if design[0 : i+1] is not a valid pattern,
            #   we are done matching characters
            if design[i] not in node.children:
                break
            # move along the pattern
            node = node.children[design[i]]
            # we reached the end of a pattern
            if node.end:
                # get the pattern combinations count for the rest of the design / sub-design
                # all of them count for this design / sub-design
                pattern_comb_count += backtrack(design[i + 1 :])

        # save the pattern combinations count for this design / sub-design
        memo[design] = pattern_comb_count
        return pattern_comb_count

    pattern_comb_counts = []
    for design in designs:
        pattern_comb_counts.append(backtrack(design))
    return pattern_comb_counts


assert sum(1 for dc in soln("sample.txt") if dc > 0) == 6
print("Part 1:", sum(1 for dc in soln("input.txt") if dc > 0))

assert sum(soln("sample.txt")) == 16
print("Part 2:", sum(soln("input.txt")))

Dart

Thanks to this useful post for reminding me that dynamic programming exists (and for linking to a source to help me remember how it works as it always makes my head spin :-) I guessed that part 2 would require counting solutions, so that helped too.

Solves live data in about 40ms.

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

int countTarget(String target, Set<String> towels) {
  int n = target.length;
  List<int> ret = List.filled(n + 1, 0)..[0] = 1;

  for (int e in 1.to(n + 1)) {
    for (int s in 0.to(e)) {
      if (towels.contains(target.substring(s, e))) ret[e] += ret[s];
    }
  }
  return ret[n];
}

List<int> allCounts(List<String> lines) {
  var towels = lines.first.split(', ').toSet();
  return lines.skip(2).map((p) => countTarget(p, towels)).toList();
}

part1(List<String> lines) => allCounts(lines).where((e) => e > 0).length;
part2(List<String> lines) => allCounts(lines).sum;

Haskell

I had several strategy switches from brute-force to pathfinding (when doing part1 input instead of example) because It simply wouldn’t finish. My solution only found the first path to the design, which is why I rewrote to only count how many towels there are for each prefix I have already built. Do that until there is either only one entry with the total combinations count or no entry and it’s impossible to build the design.

I like the final solution, its small (unlike my other solutions) and runs fast.

import Control.Arrow

import Data.Map (Map)

import qualified Data.List as List
import qualified Data.Map as Map

parse :: String -> ([String], [String])
parse = lines . init
        >>> (map (takeWhile (/= ',')) . words . head &&& drop 2)

countDesignPaths :: [String] -> String -> Map Int Int -> Int
countDesignPaths ts d es
        | Map.null es    = 0
        | ml == length d = mc
        | otherwise = countDesignPaths ts d es''
        where
                ((ml, mc), es') = Map.deleteFindMin es
                ns = List.filter (flip List.isPrefixOf (List.drop ml d))
                        >>> List.map length
                        >>> List.map (ml +)
                        $ ts
                es'' = List.foldl (\ m l' -> Map.insertWith (+) l' mc m) es'
                        $ ns
solve (ts, ds) = List.map (flip (countDesignPaths ts) (Map.singleton 0 1))
        >>> (List.length . List.filter (/= 0) &&& List.sum)
        $ ds

main = getContents
        >>= print
        . solve
        . parse