Day 2: Cube Conundrum


Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ or pastebin (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)

FAQ


🔒This post will be unlocked when there is a decent amount of submissions on the leaderboard to avoid cheating for top spots

🔓 Edit: Post has been unlocked after 6 minutes

  • mykl@lemmy.world
    link
    fedilink
    arrow-up
    3
    ·
    edit-2
    11 months ago

    Dart solution

    Quite straightforward, though there’s a sneaky trap in the test data for those of us who don’t read the rules carefully enough.

    Read, run and edit this solution in your browser: https://dartpad.dev/?id=203b3f0a9a1ad7a51daf14a1aeb6cf67

    parseLine(String s) {
      var game = s.split(': ');
      var num = int.parse(game.first.split(' ').last);
      var rounds = game.last.split('; ');
      var cubes = [
        for (var (e) in rounds)
          {
            for (var ee in e.split(', '))
              ee.split(' ').last: int.parse(ee.split(' ').first)
          }
      ];
      return MapEntry(num, cubes);
    }
    
    /// collects the max of the counts from both maps.
    Map merge2(Map a, Map b) => {
          for (var k in {...a.keys, ...b.keys}) k: max(a[k] ?? 0, b[k] ?? 0)
        };
    
    var limit = {"red": 12, "green": 13, "blue": 14};
    
    bool isGood(Map test) =>
        limit.entries.every((e) => (test[e.key] ?? 0) <= e.value);
    
    part1(List lines) => lines
        .map(parseLine)
        .where((e) => e.value.every(isGood))
        .map((e) => e.key)
        .sum;
    
    part2(List lines) => lines
        .map(parseLine)
        .map((e) => e.value.reduce(merge2))
        .map((e) => e.values.reduce((s, t) => s * t))
        .sum;
    
    • sjmulder@lemmy.sdf.org
      link
      fedilink
      arrow-up
      3
      ·
      11 months ago

      Quite straightforward, though there’s a sneaky trap in the test data for those of us who don’t read the rules carefully enough.

      What’s that? I didn’t notice anything, perhaps I was lucky.

      • mykl@lemmy.world
        link
        fedilink
        arrow-up
        3
        ·
        edit-2
        11 months ago

        Oh, I misread the rules as each game having rounds of draws without replacement and the test data gave the same result for that reading, so when I confidently submitted my answer I got a bit of a surprise.