C#

using System;
using System.Linq;

public record Point(int X, int Y); 

static class Program
{
    static async Task Main(string[] args)
    {
        var data = (await ReadInputFromFile("data.txt")).ToArray();

        var part1Answer = CalculateTotalDifference(data);
        Console.WriteLine($"Part 1 = {part1Answer}");

        var part2Answer = CountFrequencies(data);
        Console.WriteLine($"Part 2 = {part2Answer}");
    }

    public static int CountFrequencies(ICollection<Point> points)
    {
        var freq = points
            .GroupBy(p => p.Y)
            .ToDictionary(g => g.Key, g => g.Count());
        return points
            .Sum(p => freq.GetValueOrDefault(p.X, 0) * p.X);
    }

    public static int CalculateTotalDifference(ICollection<Point> points)
        => points.OrderBy(p => p.X)
            .Zip(
                points.OrderBy(p => p.Y),
                (px, py) => Math.Abs(px.X - py.Y))
            .Sum();

    public static readonly char[] Delimiter = new char[] { ' ' };
    public static async Task<IEnumerable<Point>> ReadInputFromFile(string path)
        => (await File.ReadAllLinesAsync(path))
            .Select(l =>
            {
                var parts = l.Split(
                    Delimiter,
                    StringSplitOptions.TrimEntries | StringSplitOptions.RemoveEmptyEntries);
                return new Point(int.Parse(parts[0]), int.Parse(parts[1]));
            });
}

JavaScript

After writing a procedural to-the-point version in C, tried a JavaScript solution too because it’s just perfect for list comprehension. The part 2 search is inefficient but the data size is small.

const fs = require("fs");
const U = require("./util");

const pairs = fs
    .readFileSync(process.argv[2] || process.stdin.fd, "utf8")
    .split("\n")
    .filter(x => x != "")
    .map(x => x.split(/ +/).map(Number));

const ls = pairs.map(x => x[0]); ls.sort();
const rs = pairs.map(x => x[1]); rs.sort();

const p1 = U.sum(ls.map((l, i) => Math.abs(l - rs[i])));
const p2 = U.sum(ls.map(l => l * U.count(rs, l)));

console.log("01:", p1, p2);

https://github.com/sjmulder/aoc/blob/master/2024/js/day01.js

Haskell

import Control.Arrow
import Control.Monad
import Data.List
import Data.Map

part1 [a, b] = sum $ abs <$> zipWith (-) (sort a) (sort b)
part2 [a, b] = sum $ ap (zipWith (*)) (fmap (flip (findWithDefault 0) (freq b))) a
  where
    freq = fromListWith (+) . fmap (,1)

main = getContents >>= (print . (part1 &&& part2)) . transpose . fmap (fmap read . words) . lines

TypeScript

This is for part #2 only.

import { readFileSync } from 'fs'

const f = readFileSync('./input.txt', 'utf-8')
const lines = f.split("\n")

let rights = {}
for (const i in lines) {
	if (lines[i] == '') { continue }

	const [, right] = lines[i].split(/\s+/)
	if (rights[right] === undefined) {
		rights[right] = 0
	}

	rights[right]++
}

let ans = 0
for (const i in lines) {
	const [left] = lines[i].split(/\s+/)
	const similarity = rights[left]

	if (similarity) {
		ans += (Number(left) * rights[left])
	}
}

console.dir(ans)

Is it possible to get this more efficient? I would love a way that only required iterating over the list once, but I don’t really have the focus to puzzle it out any less than O(2n) (probably more than that, even, if you count reading in the data…).

Smalltalk

day1p12: input    
	| list1 list2 nums dist sim |
	
	list1 := OrderedCollection new.
	list2 := OrderedCollection new.
	
	input linesDo: [ :l |
		nums := l substrings collect: [ :n | n asInteger ].
		list1 add: (nums at: 1).
		list2 add: (nums at: 2).
	].

	list1 sort.
	list2 sort.
	
	dist := 0.
	list1 with: list2 do: [ :a :b | dist := dist + (a - b) abs ].
	
	sim := list1 sumNumbers: [ :x | x * (list2 occurrencesOf: x) ].
	
	^ Array with: dist with: sim.

Crystal

f = ARGV[0]? ? File.read_lines("input.txt") : test.lines
list1 = Array(Int32).new(f.size)
list2 = Array(Int32).new(f.size)

f.each do |l|
	nums = l.split.map(&.to_i)
	list1.push(nums[0])
	list2.push(nums[1])
end

list1.sort!
list2.sort!

puts list1.zip(list2).sum{ |l1, l2| (l1 - l2).abs }

puts list1.sum {|x| x * list2.count x}

Python

ids = """<multiline input string>"""
ids = [pair.split(" ") for pair in ids.split("\n")]
left = [int(t[0]) for t in ids]
right = [int(t[-1]) for t in ids]
left.sort()
right.sort()
print(sum(abs(l - r) for l, r in zip(left, right)))

# 2
s = 0
for l in left:
    s += right.count(l) * l
print(s)

Lost a minute because I forgot about abs .-.

Viml

I think viml is a very fun language, i like weird languages lol, so this year im doing it in viml while trying to use as many of the original ed/ex commands as i can (:d, :p, :a, :g, …)

!cp ./puzzle1 ./puzzle1.editing
e ./puzzle1.editing

1,$sort
let row1 = []

g/^\d/let row1 = add(row1, str2nr(expand("<cword>"))) | norm 0dw
1d
1,$sort
g/^\d/execute 'norm cc' .. string(abs(expand("<cword>") - row1[line('.') - 1]))

$a|---ANSWER---
0
.
1,$-1g/^\d/call setline("$", str2nr(getline("$")) + str2nr(expand("<cword>")))

read ./puzzle1

let cnt = 0
g/^\d/let cnt += expand("<cword>") *
            \ searchcount(#{pattern: '\s\+' .. expand("<cword>")}).total

echo cnt .. "\n"

w! ./puzzle1.editing

Raku

I’m trying warm up to Raku again.

use v6;

sub MAIN($input) {
    my $file = open $input;

    grammar LocationList {
        token TOP { <row>+%"\n" "\n"* }
        token row { <left=.id> " "+ <right=.id> }
        token id { \d+ }
    }

    my $locations = LocationList.parse($file.slurp);
    my @rows = $locations<row>.map({ (.<left>.Int, .<right>.Int)});
    my $part-one-solution = (@rows[*;0].sort Z- @rows[*;1].sort)».abs.sum;
    say "part 1: $part-one-solution";

    my $rbag = bag(@rows[*;1].sort);
    my $part-two-solution = @rows[*;0].map({ $_ * $rbag{$_}}).sum;
    say "part 2: $part-two-solution";
}

I’m happy to see that Lemmy no longer eats Raku code.

Uiua

For entertainment purposes only, I’ll be trying a solution in Uiua each day until it all gets too much for me…

$ 3   4
$ 4   3
$ 2   5
$ 1   3
$ 3   9
$ 3   3
⊜∘⊸≠@\n     # Partition at \n.
⊜(⍆∵⋕)⊸≠@\s # Partition at space, parse ints, sort.

&p/+/(⌵-). # Part1 : Get abs differences, sum, print.

&p/+×⟜(/+⍉≡⌕)°⊂ # Part 2 : Count instances, mul out, sum, print.

I’m late to the party, as usual. Damned timezones. This year I’m going to tackle with a small handful of languages, but primarily Elixir and Gleam. This is my first time trying this languages in earnest, so expect some terrible, inefficient and totally unidiomatic code!
Here’s day one:

Elixir

part_one =
  File.read!("input.in")
  |> String.split("\n", trim: true)
  |> Enum.map(fn line ->
    line
    |> String.split()
    |> Enum.map(&String.to_integer/1)
  end)
  |> Enum.reduce({[], []}, fn [first, second], {list1, list2} ->
    {[first | list1], [second | list2]}
  end)
  |> then(fn {list1, list2} ->
    {Enum.sort(list1), Enum.sort(list2)}
  end)
  |> then(fn {list1, list2} ->
    Enum.zip(list1, list2)
    |> Enum.map(fn {x, y} -> abs(x - y) end)
  end)
  |> Enum.sum()

part_two =
  File.read!("input.in")
  |> String.split("\n", trim: true)
  |> Enum.map(fn line ->
    line
    |> String.split()
    |> Enum.map(&String.to_integer/1)
  end)
  |> Enum.reduce({[], []}, fn [first, second], {list1, list2} ->
    {[first | list1], [second | list2]}
  end)
  |> then(fn {list1, list2} ->
    Enum.map(list1, fn line ->
      line * Enum.count(list2, fn x -> x === line end)
    end)
    |> Enum.sum()
  end)

IO.inspect(part_one)
IO.inspect(part_two)

Haskell

Plenty of scope for making part 2 faster, but I think simple is best here. Forgot to sort the lists in the first part, which pushed me waaay off the leaderboard.

import Data.List

main = do
  [as, bs] <- transpose . map (map read . words) . lines <$> readFile "input01"
  print . sum $ map abs $ zipWith (-) (sort as) (sort bs)
  print . sum $ map (\a -> a * length (filter (== a) bs)) as

TypeScript

import { AdventOfCodeSolutionFunction } from "./solutions";

function InstancesOf(sorted_array: Array<number>, value: number) {
    const index = sorted_array.indexOf(value);
    if(index == -1)
        return 0;

    let sum = 1;

    for (let array_index = index + 1; array_index < sorted_array.length; array_index++) {
        if(sorted_array[array_index] != value)
            break;

        sum += 1;
    }

    return sum;
}

export const solution_1: AdventOfCodeSolutionFunction = (input) => {
    const left: Array<number> = [];
    const right: Array<number> = [];

    const lines = input.split("\n");

    for (let index = 0; index < lines.length; index++) {
        const element = lines[index].trim();
        if(!element)
            continue;

        const leftRight = element.split("   ");
        left.push(Number(leftRight[0]));
        right.push(Number(leftRight[1]));
    }

    const numSort = (a: number, b: number) => a - b;
    left.sort(numSort);
    right.sort(numSort);

    let sum = 0;
    for (let index = 0; index < left.length; index++) {
        const leftValue = left[index];
        const rightValue = right[index];

        sum += Math.abs(leftValue - rightValue);
    }

    const part1 = `Part 1: ${sum}`;

    sum = 0;
    for (let index = 0; index < left.length; index++) {
        sum += left[index] * InstancesOf(right, left[index]);
    }

    const part2 = `Part 2: ${sum}`;

    return `${part1}\n${part2}`;
};

I’m quite inexperienced as a programmer, I learned most of the basic concepts from playing human resource machine and 7 billion humans. After mucking about writing some CLI utilities in Perl and python, I’ve decided to give rust a go.

Part 1

Part 2

python

I didn’t realize it was december until this afternoon. I’ve generally chosen a new or spartan lang to challenge myself, but I’m going easy mode this year with python and just focusing on meeting the requirement.

import aoc

def setup():
    lines = aoc.get_lines(1)
    l = [int(x.split()[0]) for x in lines]
    r = [int(x.split()[1]) for x in lines]
    return (l, r, 0)

def one():
    l, r, acc = setup()
    for p in zip(sorted(l), sorted(r)):
        acc += abs(p[0] - p[1])
    print(acc)

def two():
    l, r, acc = setup()
    for n in l:
        acc += n * r.count(n)
    print(acc)

one()
two()