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submitted 9 months ago* (last edited 9 months ago) by Ategon@programming.dev to c/advent_of_code@programming.dev

Day 7: Camel Cards

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[-] mykl@lemmy.world 4 points 9 months ago* (last edited 9 months ago)

Dart

I'm glad I took the time to read the directions very carefully before starting coding :-)

Top Tip: my ranking of hand types relies on the fact that if you count instances of each face and sort the resulting list from high to low, you get a list that when compared with lists from other hands gives an exact correspondence with the order of the hand types as defined, so no need for a bunch of if/thens, just

  var type = Multiset.from(hand).counts.sorted(descending).join('');

Otherwise it should all be pretty self-explanatory apart from where I chose to map card rank to hex digits in order to facilitate sorting, so 'b' means 'J'!

int descending(T a, T b) => b.compareTo(a);
var cToH = "  23456789TJQKA"; // used to map card rank to hex for sorting.

handType(List hand, {wildcard = false}) {
  var type = Multiset.from(hand).counts.sorted(descending).join('');
  var i = hand.indexOf('b');
  return (!wildcard || i == -1)
      ? type
      : '23456789acde'
          .split('')
          .map((e) => handType(hand.toList()..[i] = e, wildcard: true))
          .fold(type, (s, t) => s.compareTo(t) >= 0 ? s : t);
}

solve(List lines, {wildcard = false}) => lines
    .map((e) {
      var l = e.split(' ');
      var hand =
          l.first.split('').map((e) => cToH.indexOf(e).toRadixString(16));
      var type = handType(hand.toList(), wildcard: wildcard);
      if (wildcard) hand = hand.map((e) => e == 'b' ? '0' : e);
      return (hand.join(), type, int.parse(l.last));
    })
    .sorted((a, b) {
      var c = a.$2.compareTo(b.$2);
      return (c == 0) ? a.$1.compareTo(b.$1) : c;
    })
    .indexed(offset: 1)
    .map((e) => e.value.$3 * e.index)
    .sum;

part1(List lines) => solve(lines);

part2(List lines) => solve(lines, wildcard: true);
this post was submitted on 07 Dec 2023
18 points (100.0% liked)

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