this post was submitted on 07 Dec 2023
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev!

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.

AoC 2023

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Day 7: Camel Cards

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[โ€“] pnutzh4x0r@lemmy.ndlug.org 2 points 11 months ago* (last edited 11 months ago)

Language: Python

This was fun. More enjoyable than I initially thought (though I've done card sorting code before).

Part 1

This was pretty straightforward: create a histogram of the cards in each hand to determine their type, and if there is a tie-breaker, compare each card pairwise. I use the Counter class from collections to do the counting, and then had a dictionary/table to convert labels to numeric values for comparison. I used a very OOP approach and wrote a magic method for comparing hands and used that with Python's builtin sort. I even got to use Enum!

LABELS = {l: v for v, l in enumerate('23456789TJQKA', 2)}

class HandType(IntEnum):
    FIVE_OF_A_KIND  = 6
    FOUR_OF_A_KIND  = 5
    FULL_HOUSE      = 4
    THREE_OF_A_KIND = 3
    TWO_PAIR        = 2
    ONE_PAIR        = 1
    HIGH_CARD       = 0

class Hand:
    def __init__(self, cards=str, bid=str):
        self.cards  = cards
        self.bid    = int(bid)
        counts      = Counter(self.cards)
        self.type   = (
            HandType.FIVE_OF_A_KIND  if len(counts) == 1 else
            HandType.FOUR_OF_A_KIND  if len(counts) == 2 and any(l for l, count in counts.items() if count == 4) else
            HandType.FULL_HOUSE      if len(counts) == 2 and any(l for l, count in counts.items() if count == 3) else
            HandType.THREE_OF_A_KIND if len(counts) == 3 and any(l for l, count in counts.items() if count == 3) else
            HandType.TWO_PAIR        if len(counts) == 3 and any(l for l, count in counts.items() if count == 2) else
            HandType.ONE_PAIR        if len(counts) == 4 and any(l for l, count in counts.items() if count == 2) else
            HandType.HIGH_CARD
        )

    def __lt__(self, other):
        if self.type == other.type:
            for s_label, o_label in zip(self.cards, other.cards):
                if LABELS[s_label] == LABELS[o_label]:
                    continue
                return LABELS[s_label] < LABELS[o_label]
            return False
        return self.type < other.type

    def __repr__(self):
        return f'Hand(cards={self.cards},bid={self.bid},type={self.type})'

def read_hands(stream=sys.stdin) -> list[Hand]:
    return [Hand(*line.split()) for line in stream]

def main(stream=sys.stdin) -> None:
    hands    = sorted(read_hands(stream))
    winnings = sum(rank * hand.bid for rank, hand in enumerate(hands, 1))
    print(winnings)

Part 2

For the second part, I just had to add some post-processing code to convert the jokers into actual cards. The key insight is to find the highest and most numerous non-Joker card and convert all the Jokers to that card label.

This had two edge cases that tripped me up:

  1. 'JJJJJ': There is no other non-Joker here, so I messed up and ranked this the lowest because I ended up removing all counts.

  2. 'JJJ12': This also messed me up b/c the Joker was the most numerous card, and I didn't handle that properly.

Once I fixed the post-processing code though, everything else remained the same. Below, I only show the parts that changed from Part A.

LABELS = {l: v for v, l in enumerate('J23456789TQKA', 1)}

...

class Hand:
    def __init__(self, cards=str, bid=str):
        self.cards  = cards
        self.bid    = int(bid)
        counts      = Counter(self.cards)

        if 'J' in counts and len(counts) > 1:
            max_label = max(set(counts) - {'J'}, key=lambda l: (counts[l], LABELS[l]))
            counts[max_label] += counts['J']
            del counts['J']

        self.type   = (...)

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