this post was submitted on 07 Dec 2023
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Advent Of Code

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Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

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submitted 11 months ago* (last edited 11 months ago) by Ategon to c/advent_of_code
 

Day 7: Camel Cards

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/ , pastebin, or github (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)

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[โ€“] [email protected] 3 points 11 months ago (2 children)

Sure! This generates a number for every hand, so that a better hand gets a higher number. The resulting number will contain 11 hexadecimal digits:

0x100000 bbbbb
  ^^^^^^ \____ the hand itself
  |||||\_ 1 if "one pair"
  ||||\__ 1 if "two pairs"
  |||\___ 1 if "three of a kind"
  ||\____ 1 if "full house"
  |\_____ 1 if "four of a kind"
  \______ 1 if "five of a kind"

For example:
 AAAAA: 0x100000 bbbbb
 AAAA2: 0x010000 bbbb0
 22233: 0x001000 00011

The hand itself is 5 hexadecimal digits for every card, 0 for "2" to b for "ace".

This way the higher combination always has a higher number, and hands with the same combination are ordered by the order of the cards in the hand.

[โ€“] [email protected] 2 points 11 months ago* (last edited 11 months ago) (1 children)

Wow, this is exactly what I did, but in C#. That's cool.

    public class Hand
    {
        public string Cards;
        public int Rank;
        public int Bid;
    }

    public static HandType IdentifyHandType(string hand)
    {
        var cardCounts = hand
            .Aggregate(
                new Dictionary(),
                (counts, card) => 
                {
                    counts[card] = counts.TryGetValue(card, out var count) ? (count + 1) : 1;
                    return counts;
                })
            .OrderByDescending(kvp => kvp.Value);

        using (var cardCount = cardCounts.GetEnumerator())
        {
            cardCount.MoveNext();
            switch (cardCount.Current.Value)
            {
                case 5: return HandType.FiveOfAKind;
                case 4: return HandType.FourOfAKind;
                case 3: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.FullHouse : HandType.ThreeOfAKind; }
                case 2: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.TwoPairs : HandType.OnePair; }
            }
        }

        return HandType.HighCard;
    }

    public static Hand SetHandRank(Hand hand, Dictionary cardValues)
    {
        int rank = 0;
        int offset = 0;

        var cardValueHand = hand.Cards;
        for (int i = cardValueHand.Length - 1; i >= 0; i--)
        {
            var card = cardValueHand[i];
            var cardValue = cardValues[card];
            var offsetCardValue = cardValue << offset;
            rank |= offsetCardValue;
            offset += 4; // To store values up to 13 we need 4 bits.
        }

        // Put the hand type at the high end because it is the most
        // important factor in the rank.
        var handType = (int)IdentifyHandType(hand.Cards);
        var offsetHandType = handType << offset;
        rank |= offsetHandType;

        hand.Rank = rank;
        return hand;
    }
[โ€“] snowe 2 points 11 months ago (1 children)

That is a really cool solution. Thanks for the explanation! I took a much more... um... naive path lol.

[โ€“] [email protected] 2 points 11 months ago (1 children)

I think you have the same solution, basically, just the details are a bit different. I like how you handled the joker, I didn't realise you could just multiply your best streak of cards to get the best possible combination.

[โ€“] snowe 1 points 11 months ago (1 children)

I didn't multiply the streak, I just took the jokers and added them to the highest hand already in the list. Is that not what you did? It looked the same to me.

[โ€“] [email protected] 1 points 11 months ago

This is what I meant, but I phrased it poorly :)

In my solution I reimplement the logic of identifying the hand value, but with the presence of joker (instead of just reusing the same logic).