this post was submitted on 10 Dec 2024
15 points (89.5% liked)

Advent Of Code

920 readers
58 users here now

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 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 18 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 1 year ago
MODERATORS
 

Day 10: Hoof It

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)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

you are viewing a single comment's thread
view the rest of the comments
[โ€“] [email protected] 2 points 1 week ago

Rust

This was a nice one. Basically 9 rounds of Breadth-First-Search, which could be neatly expressed using fold. The only difference between part 1 and part 2 turned out to be the datastructure for the search frontier: The HashSet in part 1 unifies paths as they join back to the same node, the Vec in part 2 keeps all paths separate.

Solution

use std::collections::HashSet;

fn parse(input: &str) -> Vec<&[u8]> {
    input.lines().map(|l| l.as_bytes()).collect()
}

fn adj(grid: &[&[u8]], (x, y): (usize, usize)) -> Vec<(usize, usize)> {
    let n = grid[y][x];
    let mut adj = Vec::with_capacity(4);
    if x > 0 && grid[y][x - 1] == n + 1 {
        adj.push((x - 1, y))
    }
    if y > 0 && grid[y - 1][x] == n + 1 {
        adj.push((x, y - 1))
    }
    if x + 1 < grid[0].len() && grid[y][x + 1] == n + 1 {
        adj.push((x + 1, y))
    }
    if y + 1 < grid.len() && grid[y + 1][x] == n + 1 {
        adj.push((x, y + 1))
    }
    adj
}

fn solve(input: String, trailhead: fn(&[&[u8]], (usize, usize)) -> u32) -> u32 {
    let grid = parse(&input);
    let mut sum = 0;
    for (y, row) in grid.iter().enumerate() {
        for (x, p) in row.iter().enumerate() {
            if *p == b'0' {
                sum += trailhead(&grid, (x, y));
            }
        }
    }
    sum
}

fn part1(input: String) {
    fn score(grid: &[&[u8]], start: (usize, usize)) -> u32 {
        (1..=9)
            .fold(HashSet::from([start]), |frontier, _| {
                frontier.iter().flat_map(|p| adj(grid, *p)).collect()
            })
            .len() as u32
    }
    println!("{}", solve(input, score))
}

fn part2(input: String) {
    fn rating(grid: &[&[u8]], start: (usize, usize)) -> u32 {
        (1..=9)
            .fold(vec![start], |frontier, _| {
                frontier.iter().flat_map(|p| adj(grid, *p)).collect()
            })
            .len() as u32
    }
    println!("{}", solve(input, rating))
}

util::aoc_main!();

Also on github