Advent of Code 2023 Day 7

Day 7: Camel Cards

https://adventofcode.com/2023/day/7

You arrive at Desert Island, an elf is there waiting to take you to your destination. A trip by camel!

The elf wants to play a game while travelling appropriately named camel cards.

It’s similar to poker, but a bit simpler, because that makes it easier to play while riding a camel.

You get a list of hands of cards. You need to order those hands based on the strength of each hand.

A hand consists of five cards.

The cards are all one of:

Each hand of 5 cards has a type of:

Five of a kind

All cards have the same label.

eg: AAAAA
Four of a kind

Four cards have the same label.

eg: AA8AA
Full House

Three cards have the same label, the two other cards have the same label.

eg: 23332
Three of a kind

Three cards have the same label.

eg: TTT98
Two pair

Two cards have the same label.
Two other cards also have the same label.

eg: 23432
One pair

Two cards have the same label.

eg: A23A4
High card

All card labels are different.

eg: 23456

Each hand has a bid associated with it, it number.

If two hands have the same type, their order is determined by looking which hand has a stronger card first.

eg: hand1 is 33332, and hand2 is 2AAAA.

Both hands have the same type (four of a kind). 33332 is stronger because the first card is stronger. If the first card is identical, the second card is compared, and so forth until one hand is stronger (two hands being completely identical never happens in the input).

Today’s input is a list of hands with their corresponding bids.

An example input looks like this:

32T3K 765
T55J5 684
KK677 28
KTJJT 220
QQQJA 483

Part 1

Order the hands in the input.

The weakest hand gets rank 1, the next one rank 2, and so on.

For each hand, multiply its rank by its bid to determine the hands winnings.

The question asks you determine the total winnings for all hands.

Option 1: lots of functions

Starting with some pseudo/skeleton-code again:

input
    .lines()
    .map(/* turn a line into a (cards, bid) pair*/)
    .sorted()
    .map(/* multiply rank with bid */)
    .sum()

Helpers

I chose to represent the value of a single card as a number, ordered like the question specifies.

fn card_value(c: char) -> usize {
    "23456789TJQKA".chars().position(|card| card == c).unwrap()
}

To score a single hand, I used a list of 2 sorted numbers:

The amount of cards that appear the most often
The amount of cards that appears the 2nd most often

Some examples:

[3, 2] represents a hand with 3 cards of equal value, and 2 cards with a different identical value
[3, 1] represents a hand with 3 cards of equal value, and 1 card with a different value. (not represented is the last card that also has a different value)
[5, 0] represents a hand with 5 cards of equal value.
[2, 2] represents a hand with 2 pairs.

fn hand_score(hand: &str) -> [u8; 2] {
    let mut faces = [0_u8; 13];
    for c in hand.chars() {
        faces[card_value(c)] += 1;
    }
    faces.sort_unstable();
    let mut score: [u8; 2] = faces[11..].try_into().unwrap();
    score.reverse();
    score
}

And finally, using those building blocks, a way to compare 2 hands and determine which one is stronger.

fn compare_hands(a: &str, b: &str) -> Ordering {
    hand_score(a).cmp(&hand_score(b)).then_with(|| {
        let (a, b) = a
            .chars()
            .zip(b.chars())
            .find(|(a, b)| a != b)
            .expect("hands are completely identical");

        card_value(a).cmp(&card_value(b))
    })
}

Code

use itertools::Itertools;
use std::cmp::Ordering;

pub fn part_1(input: &str) -> usize {
    input
        .lines()
        .map(|line| {
            let (hand, bid) = line.split_once(" ").unwrap();
            (hand, bid)
        })
        .sorted_by(|(a, _), (b, _)| compare_hands(a, b))
        .enumerate()
        .map(|(idx, (_, bid))| (idx + 1) * bid.parse::<usize>().unwrap())
        .sum()
}

fn compare_hands(a: &str, b: &str) -> Ordering {
    hand_score(a).cmp(&hand_score(b)).then_with(|| {
        let (a, b) = a
            .chars()
            .zip(b.chars())
            .find(|(a, b)| a != b)
            .expect("hands are completely identical");

        card_value(a).cmp(&card_value(b))
    })
}

fn hand_score(hand: &str) -> [u8; 2] {
    let mut faces = [0_u8; 13];
    for c in hand.chars() {
        faces[card_value(c)] += 1;
    }
    faces.sort_unstable();
    let mut score: [u8; 2] = faces[11..].try_into().unwrap();
    score.reverse();
    score
}

fn card_value(c: char) -> usize {
    "23456789TJQKA".chars().position(|card| card == c).unwrap()
}

Option 2: structures

I like to put data in neat structures. Rust structs are great.

Because the input is a string, I store references to that string in those structures. Those need a lifetime, and that’s that 'a stuff you see all over the place.

I’m not going deep into what that means in this post, but in a nutshell it means “this string is valid while this structure exists”.

By Implementing Ord we tell Rust those structures can be ordered.

Code

use itertools::Itertools;
use std::cmp::Ordering;

#[derive(PartialEq, Eq)]
struct Hand<'a> {
    cards: &'a str,
    bid: u32,
}

impl<'a> Hand<'a> {
    fn score(&self) -> [u8; 2] {
        let mut faces = [0; 13];
        for c in self.cards.chars() {
            faces[card_value(c)] += 1;
        }
        faces.sort_unstable();
        let mut score: [u8; 2] = faces[11..].try_into().unwrap();
        score.reverse();
        score
    }
}

impl<'a> PartialOrd for Hand<'a> {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl<'a> Ord for Hand<'a> {
    fn cmp(&self, other: &Self) -> Ordering {
        self.score().cmp(&other.score()).then_with(|| {
            let (a, b) = self
                .cards
                .chars()
                .zip(other.cards.chars())
                .find(|(a, b)| a != b)
                .expect("hands are completely identical");

            card_value(a).cmp(&card_value(b))
        })
    }
}

pub fn part_1(input: &str) -> u32 {
    input
        .lines()
        .map(|line| {
            let (cards, bid) = line.split_once(" ").unwrap();
            Hand {
                cards,
                bid: bid.parse().unwrap(),
            }
        })
        .sorted_unstable()
        .enumerate()
        .map(|(idx, turn)| (idx as u32 + 1) * turn.bid)
        .sum()
}

Option 3: getting shifty

I didn’t implement this but you can also represent a hand as a single number.

4 bits represent each card. The starting bits represent the type of hand (from five of a kind to high card).

This would work by determining the value of each card, and shifting bits to represent that value onto a number. Then, shift a value representing the type of the hand onto the front of the same number.

Sorting those numbers would apply the sorting logic described where it sorts based on the type of hand first, and on the individual cards next.

Part 2

There is a second way to play this game.

The J cards are now jokers, wildcards that can act like any other card.

You can use these jokers in a hand to make a stronger type of hand.

To balance this, the J card is now the weakest card.

This causes a few adjustments to the code from part1.

Helpers

The function to determine the value of a single card:

fn card_value2(c: char) -> usize {
    "J23456789TQKA".chars().position(|card| card == c).unwrap()
}

The function to score a hand:

fn hand_score2(hand: &str) -> [u8; 2] {
    let mut faces = [0; 13];
    let mut jokers = 0;
    for c in hand.chars() {
        if c == 'J' {
            jokers += 1;
        } else {
            faces[card_value2(c)] += 1;
        }
    }
    faces.sort_unstable();
    let mut score: [u8; 2] = faces[11..].try_into().unwrap();
    score.reverse();
    // add the amount of jokers to the counts of the card that occurs the most already to increase the hand score
    score[0] += jokers;
    score
}

The function to compare 2 hands:

fn compare_hands2(a: &str, b: &str) -> Ordering {
    hand_score2(a).cmp(&hand_score2(b)).then_with(|| {
        let (a, b) = a
            .chars()
            .zip(b.chars())
            .find(|(a, b)| a != b)
            .expect("hands are completely identical");

        card_value2(a).cmp(&card_value2(b))
    })
}

The result is nearly identical code to part one. It would be trivial to implement a boolean flag and combine these 2 version into one, but I had somewhere to go and copy-pasting was faster.

Code

use itertools::Itertools;
use std::cmp::Ordering;

pub fn part_2(input: &str) -> usize {
    input
        .lines()
        .map(|line| {
            let (hand, bid) = line.split_once(" ").unwrap();
            (hand, bid)
        })
        .sorted_by(|(a, _), (b, _)| compare_hands2(a, b))
        .enumerate()
        .map(|(idx, (_, bid))| (idx + 1) * bid.parse::<usize>().unwrap())
        .sum()
}

fn compare_hands2(a: &str, b: &str) -> Ordering {
    hand_score2(a).cmp(&hand_score2(b)).then_with(|| {
        let (a, b) = a
            .chars()
            .zip(b.chars())
            .find(|(a, b)| a != b)
            .expect("hands are completely identical");

        card_value2(a).cmp(&card_value2(b))
    })
}

fn hand_score2(hand: &str) -> [u8; 2] {
    let mut faces = [0; 13];
    let mut jokers = 0;
    for c in hand.chars() {
        if c == 'J' {
            jokers += 1;
        } else {
            faces[card_value2(c)] += 1;
        }
    }
    faces.sort_unstable();
    let mut score: [u8; 2] = faces[11..].try_into().unwrap();
    score.reverse();
    // add the amount of jokers to the counts of the card that occurs the most already to increase the hand score
    score[0] += jokers;
    score
}

fn card_value2(c: char) -> usize {
    "J23456789TQKA".chars().position(|card| card == c).unwrap()
}

Final code

use itertools::Itertools;
use std::cmp::Ordering;

#[derive(PartialEq, Eq)]
struct Hand<'a> {
    cards: &'a str,
    bid: u32,
}

impl<'a> Hand<'a> {
    fn score(&self) -> [u8; 2] {
        let mut faces = [0; 13];
        for c in self.cards.chars() {
            faces[card_value(c)] += 1;
        }
        faces.sort_unstable();
        let mut score: [u8; 2] = faces[11..].try_into().unwrap();
        score.reverse();
        score
    }
}

impl<'a> PartialOrd for Hand<'a> {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl<'a> Ord for Hand<'a> {
    fn cmp(&self, other: &Self) -> Ordering {
        self.score().cmp(&other.score()).then_with(|| {
            let (a, b) = self
                .cards
                .chars()
                .zip(other.cards.chars())
                .find(|(a, b)| a != b)
                .expect("hands are completely identical");

            card_value(a).cmp(&card_value(b))
        })
    }
}

pub fn part_1(input: &str) -> u32 {
    input
        .lines()
        .map(|line| {
            let (cards, bid) = line.split_once(" ").unwrap();
            Hand {
                cards,
                bid: bid.parse().unwrap(),
            }
        })
        .sorted_unstable()
        .enumerate()
        .map(|(idx, turn)| (idx as u32 + 1) * turn.bid)
        .sum()
}

#[derive(PartialEq, Eq)]
struct Hand2<'a> {
    cards: &'a str,
    bid: u32,
}

impl<'a> Hand2<'a> {
    fn score(&self) -> [u8; 2] {
        let mut faces = [0; 13];
        let mut jokers = 0;
        for c in self.cards.chars() {
            if c == 'J' {
                jokers += 1;
            } else {
                faces[card_value2(c)] += 1;
            }
        }
        faces.sort_unstable();
        let mut score: [u8; 2] = faces[11..].try_into().unwrap();
        score.reverse();
        // add the amount of jokers to the counts of the card that occurs the most already to increase the hand score
        score[0] += jokers;
        score
    }
}

impl<'a> PartialOrd for Hand2<'a> {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl<'a> Ord for Hand2<'a> {
    fn cmp(&self, other: &Self) -> Ordering {
        self.score().cmp(&other.score()).then_with(|| {
            let (a, b) = self
                .cards
                .chars()
                .zip(other.cards.chars())
                .find(|(a, b)| a != b)
                .expect("hands are completely identical");

            card_value2(a).cmp(&card_value2(b))
        })
    }
}

pub fn part_2(input: &str) -> u32 {
    input
        .lines()
        .map(|line| {
            let (cards, bid) = line.split_once(" ").unwrap();
            Hand2 {
                cards,
                bid: bid.parse().unwrap(),
            }
        })
        .sorted_unstable()
        .enumerate()
        .map(|(idx, turn)| (idx as u32 + 1) * turn.bid)
        .sum()
}

fn card_value(c: char) -> usize {
    "23456789TJQKA".chars().position(|card| card == c).unwrap()
}

fn card_value2(c: char) -> usize {
    "J23456789TQKA".chars().position(|card| card == c).unwrap()
}

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Advent of Code 2023 Day 7

Day 7: Camel Cards

Part 1

Option 1: lots of functions

Helpers

Code

Option 2: structures

Code

Option 3: getting shifty

Part 2

Helpers

Code

Final code