Python

After a bunch of fiddling yesterday and today I finally managed to arrive at a regex-only solution for part 2. That re.DOTALL is crucial here.

import re
from pathlib import Path


def parse_input_one(input: str) -> list[tuple[int]]:
    p = re.compile(r"mul\((\d{1,3}),(\d{1,3})\)")
    return [(int(m[0]), int(m[1])) for m in p.findall(input)]


def parse_input_two(input: str) -> list[tuple[int]]:
    p = re.compile(r"don't\(\).*?do\(\)|mul\((\d{1,3}),(\d{1,3})\)", re.DOTALL)
    return [(int(m[0]), int(m[1])) for m in p.findall(input) if m[0] and m[1]]


def part_one(input: str) -> int:
    pairs = parse_input_one(input)
    return sum(map(lambda v: v[0] * v[1], pairs))


def part_two(input: str) -> int:
    pairs = parse_input_two(input)
    return sum(map(lambda v: v[0] * v[1], pairs))


if __name__ == "__main__":
    input = Path("input").read_text("utf-8")
    print(part_one(input))
    print(part_two(input))

Nim

import ../aoc, re, sequtils, strutils, math

proc mulsum*(line:string):int=
  let matches = line.findAll(re"mul\([0-9]{1,3},[0-9]{1,3}\)")
  let pairs = matches.mapIt(it[4..^2].split(',').map(parseInt))
  pairs.mapIt(it[0]*it[1]).sum

proc filter*(line:string):int=
  var state = true;
  var i=0
  while i < line.len:
    if state:
      let off = line.find("don't()", i)
      if off == -1:
        break
      result += line[i..<off].mulsum
      i = off+6
      state = false
    else:
      let on = line.find("do()", i)
      if on == -1:
        break
      i = on+4
      state = true
      
  if state:
    result += line[i..^1].mulsum

proc solve*(input:string): array[2,int] =
  #part 1&2
  result = [input.mulsum, input.filter]

I had a nicer solution in mind for part 2, but for some reason nre didn't want to work for me, and re couldn't give me the start/end or all results, so I ended up doing this skip/toggle approach.

Also initially I was doing it line by line out of habit from other puzzles, but then ofc the don't()s didn't propagate to the next line.

Python

I'm surprised I don't see more people taking advantage of eval I thought it was pretty slick.

import operator 
import re

with open('input.txt', 'r') as file:
    memory = file.read()

matches = re.findall("mul\(\d{1,3},\d{1,3}\)|don't\(\)|do\(\)", memory)

enabled = 1
filtered_matches = []
for instruction in matches:
    if instruction == "don't()":
        enabled = 0
        continue
    elif instruction == "do()":
        enabled = 1
        continue
    elif enabled: 
        filtered_matches.append(instruction)
multipled = [eval(f"operator.{x}") for x in filtered_matches]
print(sum(multiples))

Kotlin

Just the standard Regex stuff. I found this website to be very helpful to write the patterns. (Very useful in general)

fun main() {
    fun part1(input: List<String>): Int =
        Regex("""mul\(\d+,\d+\)""").findAll(input.joinToString()).sumOf {
            with(Regex("""\d+""").findAll(it.value)) { this.first().value.toInt() * this.last().value.toInt() }
        }

    fun part2(input: List<String>): Int {
        var isMultiplyInstructionEnabled = true  // by default
        return Regex("""mul\(\d+,\d+\)|do\(\)|don't\(\)""").findAll(input.joinToString()).fold(0) { acc, instruction ->
            when (instruction.value) {
                "do()" -> acc.also { isMultiplyInstructionEnabled = true }
                "don't()" -> acc.also { isMultiplyInstructionEnabled = false }
                else -> {
                    if (isMultiplyInstructionEnabled) {
                        acc + with(Regex("""\d+""").findAll(instruction.value)) { this.first().value.toInt() * this.last().value.toInt() }
                    } else acc
                }
            }
        }
    }

    val testInputPart1 = readInput("Day03_test_part1")
    val testInputPart2 = readInput("Day03_test_part2")
    check(part1(testInputPart1) == 161)
    check(part2(testInputPart2) == 48)

    val input = readInput("Day03")
    part1(input).println()
    part2(input).println()
}

´´´

Rust

Didn't do anything crazy here -- ended up using regex like a bunch of other folks.

use regex::Regex;

use crate::shared::util::read_lines;

fn parse_mul(input: &[String]) -> (u32, u32) {
    // Lazy, but rejoin after having removed `\n`ewlines.
    let joined = input.concat();
    let re = Regex::new(r"mul\((\d+,\d+)\)|(do\(\))|(don't\(\))").expect("invalid regex");

    // part1
    let mut total1 = 0u32;
    // part2 -- adds `do()`s and `don't()`s
    let mut total2 = 0u32;
    let mut enabled = 1u32;

    re.captures_iter(&joined).for_each(|c| {
        let (_, [m]) = c.extract();
        match m {
            "do()" => enabled = 1,
            "don't()" => enabled = 0,
            _ => {
                let product: u32 = m.split(",").map(|s| s.parse::<u32>().unwrap()).product();
                total1 += product;
                total2 += product * enabled;
            }
        }
    });
    (total1, total2)
}

pub fn solve() {
    let input = read_lines("inputs/day03.txt");
    let (part1_res, part2_res) = parse_mul(&input);
    println!("Part 1: {}", part1_res);
    println!("Part 2: {}", part2_res);
}

#[cfg(test)]
mod test {
    use super::*;

    #[test]
    fn test_solution() {
        let test_input = vec![
            "xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))".to_string(),
        ];
        let (p1, p2) = parse_mul(&test_input);
        eprintln!("P1: {p1}, P2: {p2}");
        assert_eq!(161, p1);
        assert_eq!(48, p2);
    }
}

Solution on my github (Made it public now)

Gleam

Struggled with the second part as I am still very new to this very cool language, but got there after scrolling for some inspiration.

import gleam/int
import gleam/io
import gleam/list
import gleam/regex
import gleam/result
import gleam/string
import simplifile

pub fn main() {
  let assert Ok(data) = simplifile.read("input.in")
  part_one(data) |> io.debug
  part_two(data) |> io.debug
}

fn part_one(data) {
  let assert Ok(multiplication_pattern) =
    regex.from_string("mul\\(\\d{1,3},\\d{1,3}\\)")
  let assert Ok(digit_pattern) = regex.from_string("\\d{1,3},\\d{1,3}")
  let multiplications =
    regex.scan(multiplication_pattern, data)
    |> list.flat_map(fn(reg) {
      regex.scan(digit_pattern, reg.content)
      |> list.map(fn(digits) {
        digits.content
        |> string.split(",")
        |> list.map(fn(x) { x |> int.parse |> result.unwrap(0) })
        |> list.reduce(fn(a, b) { a * b })
        |> result.unwrap(0)
      })
    })
    |> list.reduce(fn(a, b) { a + b })
    |> result.unwrap(0)
}

fn part_two(data) {
  let data = "do()" <> string.replace(data, "\n", "") <> "don't()"
  let assert Ok(pattern) = regex.from_string("do\\(\\).*?don't\\(\\)")
  regex.scan(pattern, data)
  |> list.map(fn(input) { input.content |> part_one })
  |> list.reduce(fn(a, b) { a + b })
}

Python

def process(input, part2=False):
    if part2:
        input = re.sub(r'don\'t\(\).+?do\(\)', '', input) # remove everything between don't() and do()
    total = [ int(i[0]) * int(i[1]) for i in re.findall(r'mul\((\d+),(\d+)\)', input) ]
    return sum(total)

Given the structure of the input file, we just have to ignore everything between don't() and do(), so remove those from the instructions before processing.

Python

Part1:

matches = re.findall(r"(mul\((\d+),(\d+)\))", input)
muls = [int(m[1]) * int(m[2]) for m in matches]
print(sum(muls))

Part2:

instructions = list(re.findall(r"(do\(\)|don't\(\)|(mul\((\d+),(\d+)\)))", input)
mul_enabled = True
muls = 0

for inst in instructions:
    if inst[0] == "don't()":
        mul_enabled = False
    elif inst[0] == "do()":
        mul_enabled = True
    elif mul_enabled:
        muls += int(inst[2]) * int(inst[3])

print(muls)

Uiua

Part 1:

&fras "day3/input.txt"
/+≡/×≡⋕≡↘1regex "mul\\((\\d+),(\\d+)\\)"

Part 2:

Filter ← ⍜⊜∘≡⋅""⊸⦷°□
.&fras "day3/input.txt"
∧Filter♭regex"don't\\(\\)?(.*?)(?:do\\(\\)|$)"
/+≡/×≡⋕≡↘1regex "mul\\((\\d+),(\\d+)\\)"

Raku

sub MAIN($input) {
    grammar Muls {
        token TOP { .*? <mul>+%.*? .* }
        token mul { "mul(" <number> "," <number> ")" }
        token number { \d+ }
    }

    my $parsedMuls = Muls.parsefile($input);
    my @muls = $parsedMuls<mul>.map({.<number>».Int});
    my $part-one-solution = @muls.map({[*] $_.List}).sum;
    say "part 1: $part-one-solution";

    grammar EnabledMuls {
        token TOP { .*? [<.disabled> || <mul>]+%.*? .* }
        token mul { "mul(" <number> "," <number> ")" }
        token number { \d+ }
        token disabled { "don't()" .*? ["do()" || $] }
    }

    my $parsedEnabledMuls = EnabledMuls.parsefile($input);
    my @enabledMuls = $parsedEnabledMuls<mul>.map({.<number>».Int});
    my $part-two-solution = @enabledMuls.map({[*] $_.List}).sum;
    say "part 2: $part-two-solution";
}

github

Factor

: get-input ( -- corrupted-input )
  "aoc-2024.03" "input.txt" vocab-file-path utf8 file-contents ;

: get-muls ( corrupted-input -- instructions )
  R/ mul\(\d+,\d+\)/ all-matching-subseqs ;

: process-mul ( instruction -- n )
  R/ \d+/ all-matching-subseqs
  [ string>number ] map-product ;

: solve ( corrupted-input -- n )
  get-muls [ process-mul ] map-sum ;

: part1 ( -- n )
  get-input solve ;

: part2 ( -- n )
  get-input
  R/ don't\(\)(.|\n)*?do\(\)/ split concat
  R/ don't\(\)(.|\n)*/ "" re-replace
  solve ;

Elixir

First time writing Elixir. It's probably janky af.

I've had some help from AI to get some pointers along the way. I'm not competing in any way, just trying to learn and have fun.

Part 2 is currently not working, and I can't figure out why. I'm trying to just remove everything from "don't()" to "do()" and just pass the rest through the working solution for part 1. Should work, right?

Any pointers?

defmodule Three do
  def get_input do
    File.read!("./input.txt")
  end

  def extract_operations(input) do
    Regex.scan(~r/mul\((\d{1,3}),(\d{1,3})\)/, input)
    |> Enum.map(fn [full_match, num1, num2] ->
      num1 = String.to_integer(num1)
      num2 = String.to_integer(num2)
      [full_match, num1 * num2]
    end)
  end

  def sum_products(ops) do
    ops
    |> List.flatten()
    |> Enum.filter(fn x -> is_integer(x) end)
    |> Enum.sum()
  end

  def part1 do
    extract_operations(get_input())
    |> sum_products()
  end

  def part2 do
    String.split(get_input(), ~r/don\'t\(\).*?do\(\)/)
    |> Enum.map(&extract_operations/1)
    |> sum_products()
  end
end

IO.puts("part 1: #{Three.part1()}")
IO.puts("part 2: #{Three.part2()}")

J

We can take advantage of the manageable size of the input to avoid explicit looping and mutable state; instead, construct vectors which give, for each character position in the input, the position of the most recent do() and most recent don't(); for part 2 a multiplication is enabled if the position of the most recent do() (counting start of input as 0) is greater than that of the most recent don't() (counting start of input as minus infinity).

load 'regex'

raw =: fread '3.data'
mul_matches =: 'mul\(([[:digit:]]{1,3}),([[:digit:]]{1,3})\)' rxmatches raw

NB. a b sublist y gives elements [a..b) of y
sublist =: ({~(+i.)/)~"1 _

NB. ". is number parsing
mul_results =: */"1 ". (}."2 mul_matches) sublist raw
result1 =: +/ mul_results

do_matches =: 'do\(\)' rxmatches raw
dont_matches =: 'don''t\(\)' rxmatches raw
match_indices =: (<0 0) & {"2
do_indices =: 0 , match_indices do_matches  NB. start in do mode
dont_indices =: match_indices dont_matches
NB. take successive diffs, then append length from last index to end of string
run_lengths =: (}. - }:) , (((#raw) & -) @: {:)
do_map =: (run_lengths do_indices) # do_indices
dont_map =: (({. , run_lengths) dont_indices) # __ , dont_indices
enabled =: do_map > dont_map
result2 =: +/ ((match_indices mul_matches) { enabled) * mul_results

Go

Part 1, just find the regex groups, parse to int, and done.

func part1() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	re := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
	product := 0

	for scanner.Scan() {
		line := scanner.Text()
		submatches := re.FindAllStringSubmatch(line, -1)

		for _, s := range submatches {
			a, _ := strconv.Atoi(s[1])
			b, _ := strconv.Atoi(s[2])
			product += (a * b)
		}
	}

	fmt.Println(product)
}

Part 2, not so simple. Ended up doing some weird hack with a map to check if the multiplication was enabled or not. Also instead of finding regex groups I had to find the indices, and then interpret what those mean... Not very readable code I'm afraid

func part2() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	mulRE := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
	doRE := regexp.MustCompile(`do\(\)`)
	dontRE := regexp.MustCompile(`don't\(\)`)
	product := 0
	enabled := true

	for scanner.Scan() {
		line := scanner.Text()
		doIndices := doRE.FindAllStringIndex(line, -1)
		dontIndices := dontRE.FindAllStringIndex(line, -1)
		mulSubIndices := mulRE.FindAllStringSubmatchIndex(line, -1)

		mapIndices := make(map[int]string)
		for _, do := range doIndices {
			mapIndices[do[0]] = "do"
		}
		for _, dont := range dontIndices {
			mapIndices[dont[0]] = "dont"
		}
		for _, mul := range mulSubIndices {
			mapIndices[mul[0]] = "mul"
		}

		nextMatch := 0

		for i := 0; i < len(line); i++ {
			val, ok := mapIndices[i]
			if ok && val == "do" {
				enabled = true
			} else if ok && val == "dont" {
				enabled = false
			} else if ok && val == "mul" {
				if enabled {
					match := mulSubIndices[nextMatch]
					a, _ := strconv.Atoi(string(line[match[2]:match[3]]))
					b, _ := strconv.Atoi(string(line[match[4]:match[5]]))
					product += (a * b)
				}
				nextMatch++
			}
		}
	}

	fmt.Println(product)
}

Rust

use crate::utils::read_lines;

pub fn solution1() {
    let lines = read_lines("src/day3/input.txt");
    let sum = lines
        .map(|line| {
            let mut sum = 0;
            let mut command_bytes = Vec::new();
            for byte in line.bytes() {
                match (byte, command_bytes.as_slice()) {
                    (b')', [.., b'0'..=b'9']) => {
                        handle_mul(&mut command_bytes, &mut sum);
                    }
                    _ if matches_mul(byte, &command_bytes) => {
                        command_bytes.push(byte);
                    }
                    _ => {
                        command_bytes.clear();
                    }
                }
            }

            sum
        })
        .sum::<usize>();

    println!("Sum of multiplication results = {sum}");
}

pub fn solution2() {
    let lines = read_lines("src/day3/input.txt");

    let mut can_mul = true;
    let sum = lines
        .map(|line| {
            let mut sum = 0;
            let mut command_bytes = Vec::new();
            for byte in line.bytes() {
                match (byte, command_bytes.as_slice()) {
                    (b')', [.., b'0'..=b'9']) if can_mul => {
                        handle_mul(&mut command_bytes, &mut sum);
                    }
                    (b')', [b'd', b'o', b'(']) => {
                        can_mul = true;
                        command_bytes.clear();
                    }
                    (b')', [.., b't', b'(']) => {
                        can_mul = false;
                        command_bytes.clear();
                    }
                    _ if matches_do_or_dont(byte, &command_bytes)
                        || matches_mul(byte, &command_bytes) =>
                    {
                        command_bytes.push(byte);
                    }
                    _ => {
                        command_bytes.clear();
                    }
                }
            }

            sum
        })
        .sum::<usize>();

    println!("Sum of enabled multiplication results = {sum}");
}

fn matches_mul(byte: u8, command_bytes: &[u8]) -> bool {
    matches!(
        (byte, command_bytes),
        (b'm', [])
            | (b'u', [.., b'm'])
            | (b'l', [.., b'u'])
            | (b'(', [.., b'l'])
            | (b'0'..=b'9', [.., b'(' | b'0'..=b'9' | b','])
            | (b',', [.., b'0'..=b'9'])
    )
}

fn matches_do_or_dont(byte: u8, command_bytes: &[u8]) -> bool {
    matches!(
        (byte, command_bytes),
        (b'd', [])
            | (b'o', [.., b'd'])
            | (b'n', [.., b'o'])
            | (b'\'', [.., b'n'])
            | (b'(', [.., b'o' | b't'])
            | (b't', [.., b'\''])
    )
}

fn handle_mul(command_bytes: &mut Vec<u8>, sum: &mut usize) {
    let first_num_index = command_bytes
        .iter()
        .position(u8::is_ascii_digit)
        .expect("Guarunteed to be there");
    let comma_index = command_bytes
        .iter()
        .position(|&c| c == b',')
        .expect("Guarunteed to be there.");

    let num1 = bytes_to_num(&command_bytes[first_num_index..comma_index]);
    let num2 = bytes_to_num(&command_bytes[comma_index + 1..]);

    *sum += num1 * num2;
    command_bytes.clear();
}

fn bytes_to_num(bytes: &[u8]) -> usize {
    bytes
        .iter()
        .rev()
        .enumerate()
        .map(|(i, digit)| (*digit - b'0') as usize * 10usize.pow(i as u32))
        .sum::<usize>()
}

Definitely not my prettiest code ever. It would probably look nicer if I used regex or some parsing library, but I took on the self-imposed challenge of not using third party libraries. Also, this is already further than I made it last year!

Rust with nom parser

Decided to give it a go with the nom parser (first time using this crate). Turned out quite nicely. Had some issues with the alt combinator: All alternatives have to return the same type, using a enum to wrap all options did the trick.

use memmap2::Mmap;
use nom::{
    branch::alt, bytes::complete::*, character::complete::*, combinator::map, multi::many_till,
    sequence::tuple, AsBytes, IResult,
};

#[derive(Debug)]
enum Token {
    Do,
    Dont,
    Mul(u64, u64),
}

fn main() -> anyhow::Result<()> {
    let file = std::fs::File::open("input.txt")?;
    let mmap = unsafe { Mmap::map(&file)? };

    let mut sum_part1 = 0;
    let mut sum_part2 = 0;
    let mut enabled = true;

    let mut cursor = mmap.as_bytes();
    while let Ok(token) = parse(cursor) {
        match token.1 .1 {
            Token::Do => enabled = true,
            Token::Dont => enabled = false,
            Token::Mul(left, right) => {
                let prod = left * right;
                sum_part1 += prod;
                if enabled {
                    sum_part2 += prod;
                }
            }
        }

        cursor = token.0;
    }

    println!("part1: {} part2: {}", sum_part1, sum_part2);

    Ok(())
}

type ParseResult<'a> =
    Result<(&'a [u8], (Vec<char>, Token)), nom::Err<nom::error::Error<&'a [u8]>>>;

fn parse(input: &[u8]) -> ParseResult {
    many_till(
        anychar,
        alt((
            map(doit, |_| Token::Do),
            map(dont, |_| Token::Dont),
            map(mul, |el| Token::Mul(el.2, el.4)),
        )),
    )(input)
}

fn doit(input: &[u8]) -> IResult<&[u8], &[u8]> {
    tag("do()")(input)
}

fn dont(input: &[u8]) -> IResult<&[u8], &[u8]> {
    tag("don't()")(input)
}

type ParsedMulResult<'a> = (&'a [u8], &'a [u8], u64, &'a [u8], u64, &'a [u8]);

fn mul(input: &[u8]) -> IResult<&[u8], ParsedMulResult> {
    tuple((tag("mul"), tag("("), u64, tag(","), u64, tag(")")))(input)
}

Rust

Regex made this one pretty straightforward. The second part additionally looks for do() and don't() in the same regex, then we do a case distinction on the match.

use regex::{Regex, Captures};

fn mul_cap(cap: Captures) -> i32 {
    let a = cap.get(1).unwrap().as_str().parse::<i32>().unwrap();
    let b = cap.get(2).unwrap().as_str().parse::<i32>().unwrap();
    a * b
}

fn part1(input: String) {
    let re = Regex::new(r"mul\((\d{1,3}),(\d{1,3})\)").unwrap();
    let res = re.captures_iter(&input).map(mul_cap).sum::<i32>();
    println!("{res}");
}

fn part2(input: String) {
    let re = Regex::new(r"do\(\)|don't\(\)|mul\((\d{1,3}),(\d{1,3})\)").unwrap();
    let mut enabled = true;
    let mut res = 0;
    for cap in re.captures_iter(&input) {
        match cap.get(0).unwrap().as_str() {
            "do()" => enabled = true,
            "don't()" => enabled = false,
            _ if enabled => res += mul_cap(cap),
            _ => {}
        }
    }
    println!("{res}");
}

util::aoc_main!();

Kotlin

fun part1(input: String): Int {
    val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)".toRegex()
    var sum = 0
    pattern.findAll(input).forEach { match ->
        val first = match.groups[1]?.value?.toInt()!!
        val second = match.groups[2]?.value?.toInt()!!
        sum += first * second

    }
    return sum
}

fun part2(input: String): Int {
    val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)|don't\\(\\)|do\\(\\)".toRegex()
    var sum = 0
    var enabled = true
    pattern.findAll(input).forEach { match ->
        if (match.value == "do()") enabled = true
        else if (match.value == "don't()") enabled = false
        else if (enabled) {
            val first = match.groups[1]?.value?.toInt()!!
            val second = match.groups[2]?.value?.toInt()!!
            sum += first * second
        }
    }
    return sum
}

I couldn't figure it out in haskell, so I went with bash for the first part

Shell

cat example | grep -Eo "mul\([[:digit:]]{1,3},[[:digit:]]{1,3}\)" | cut -d "(" -f 2 | tr -d ")" | tr "," "*" | paste -sd+ | bc

but this wouldn't rock anymore in the second part, so I had to resort to python for it

Python

import sys

f = "\n".join(sys.stdin.readlines())

f = f.replace("don't()", "\ndon't()\n")
f = f.replace("do()", "\ndo()\n")

import re

enabled = True
muls = []
for line in f.split("\n"):
    if line == "don't()":
        enabled = False
    if line == "do()":
        enabled = True
    if enabled:
        for match in re.finditer(r"mul\((\d{1,3}),(\d{1,3})\)", line):
            muls.append(int(match.group(1)) * int(match.group(2)))
        pass
    pass

print(sum(muls))

I started poking at doing a proper lexer/parser, but then I thought about how early in AoC it is and how low the chance is that the second part will require proper parsing.

So therefore; hello regex my old friend, I've come to talk with you again.

List<string> instructions = new List<string>();

public void Input(IEnumerable<string> lines)
{
  foreach (var line in lines)
    instructions.AddRange(Regex.Matches(line, @"mul\(\d+,\d+\)|do\(\)|don't\(\)").Select(m => m.Value));
}

public void Part1()
{
  var sum = instructions.Select(mul => Regex.Match(mul, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value))).Select(cc => cc.Aggregate(1, (acc, val) => acc * val)).Sum();
  Console.WriteLine($"Sum: {sum}");
}
public void Part2()
{
  bool enabled = true;
  long sum = 0;
  foreach(var inst in instructions)
  {
    if (inst.StartsWith("don't"))
      enabled = false;
    else if (inst.StartsWith("do"))
      enabled = true;
    else if (enabled)
      sum += Regex.Match(inst, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value)).Aggregate(1, (acc, val) => acc * val);
  }
  Console.WriteLine($"Sum: {sum}");
}

Haskell

module Main where

import Control.Arrow hiding ((+++))
import Data.Char
import Data.Functor
import Data.Maybe
import Text.ParserCombinators.ReadP hiding (get)
import Text.ParserCombinators.ReadP qualified as P

data Op = Mul Int Int | Do | Dont deriving (Show)

parser1 :: ReadP [(Int, Int)]
parser1 = catMaybes <$> many ((Just <$> mul) <++ (P.get $> Nothing))

parser2 :: ReadP [Op]
parser2 = catMaybes <$> many ((Just <$> operation) <++ (P.get $> Nothing))

mul :: ReadP (Int, Int)
mul = (,) <$> (string "mul(" *> (read <$> munch1 isDigit <* char ',')) <*> (read <$> munch1 isDigit <* char ')')

operation :: ReadP Op
operation = (string "do()" $> Do) +++ (string "don't()" $> Dont) +++ (uncurry Mul <$> mul)

foldOp :: (Bool, Int) -> Op -> (Bool, Int)
foldOp (_, n) Do = (True, n)
foldOp (_, n) Dont = (False, n)
foldOp (True, n) (Mul a b) = (True, n + a * b)
foldOp (False, n) _ = (False, n)

part1 = sum . fmap (uncurry (*)) . fst . last . readP_to_S parser1
part2 = snd . foldl foldOp (True, 0) . fst . last . readP_to_S parser2

main = getContents >>= print . (part1 &&& part2)

Uiua

Uses (abuses?) experimental features of fold to track the running state of do/don't state. Although it works well and fast, I don't think I would recommend this code to anyone :-) Try it if you must!

# Experimental!
DataP₁       ← $ xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5))
DataP₂       ← $ xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
GetMul       ← $ mul\((\d{1,3}),(\d{1,3})\)
GetMulDoDont ← $ mul\(\d{1,3},\d{1,3}\)|do\(\)|don\'t\(\)

&p/+≡(/×≡⋕↘1)regex GetMul DataP₁ # Part 1

# Build an accumulator to track running state of do/don't
Filter ← ≡⋕↘1◌∧(⍣("0" °"n"|"1" °"("|.◌)) :"1"∵◇⊡2

regex GetMulDoDont DataP₂
▽⊸≡(¬≍"do()"°□⊢)▽⊸Filter     # Apply Filter, remove the spare 'do's
&p/+≡(/×≡⋕↘1⊢regexGetMul°□⊢) # Get the digits and multiply, sum.

C

Yay parsers! I've gotten quite comfortable writing these with C. Using out pointers arguments for the cursor that are only updated if the match is successful makes for easy bookkeeping.

::: spoiler Code

#include "common.h"

static int
parse_exact(const char **stringp, const char *expect)
{
	const char *s = *stringp;
	int i;

	for (i=0; s[i] && expect[i] && s[i] == expect[i]; i++)
		;
	if (expect[i])
		return 0;

	*stringp  = &s[i];
	return 1;
}

static int
parse_int(const char **stringp, int *outp)
{
	char *end;
	int val;

	val = (int)strtol(*stringp, &end, 10);
	if (end == *stringp)
		return 0;

	*stringp = end;
	if (outp) *outp = val;
	return 1;
}

static int
parse_mul(const char **stringp, int *ap, int *bp)
{
	const char *cur = *stringp;
	int a,b;

	if (!parse_exact(&cur, "mul(") ||
	    !parse_int(&cur, &a) ||
	    !parse_exact(&cur, ",") ||
	    !parse_int(&cur, &b) ||
	    !parse_exact(&cur, ")"))
		return 0;

	*stringp = cur;
	if (ap) *ap = a;
	if (bp) *bp = b;
	return 1;
}

int
main(int argc, char **argv)
{
	static char buf[32*1024];
	const char *cur;
	size_t nr;
	int p1=0,p2=0, a,b, dont=0;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	nr = fread(buf, 1, sizeof(buf), stdin);
	assert(!ferror(stdin));
	assert(nr != sizeof(buf));
	buf[nr] = '\0';

	for (cur = buf; *cur; )
		if (parse_exact(&cur, "do()"))
			dont = 0;
		else if (parse_exact(&cur, "don't()"))
			dont = 1;
		else if (parse_mul(&cur, &a, &b)) {
			p1 += a * b;
			if (!dont) p2 += a * b;
		} else
			cur++;

	printf("03: %d %d\n", p1, p2);
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day03.c

Nim

From a first glance it was obviously a regex problem.
I'm using tinyre here instead of stdlib re library just because I'm more familiar with it.

import pkg/tinyre

proc solve(input: string): AOCSolution[int, int] =
  var allow = true
  for match in input.match(reG"mul\(\d+,\d+\)|do\(\)|don't\(\)"):
    if match == "do()": allow = true
    elif match == "don't()": allow = false
    else:
      let pair = match[4..^2].split(',')
      let mult = pair[0].parseInt * pair[1].parseInt
      result.part1 += mult
      if allow: result.part2 += mult

Codeberg repo

Haskell

Oof, a parsing problem :/ This is some nasty-ass code. step is almost the State monad written out explicitly.

import Control.Monad
import Data.Either
import Data.List
import Text.Parsec

data Ins = Mul !Int !Int | Do | Dont

readInput :: String -> [Ins]
readInput = fromRight undefined . parse input ""
  where
    input = many ins <* many anyChar
    ins =
      choice . map try $
        [ Mul <$> (string "mul(" *> arg) <*> (char ',' *> arg) <* char ')',
          Do <$ string "do()",
          Dont <$ string "don't()",
          anyChar *> ins
        ]
    arg = do
      s <- many1 digit
      guard $ length s <= 3
      return $ read s

run f = snd . foldl' step (True, 0)
  where
    step (e, a) i =
      case i of
        Mul x y -> (e, if f e then a + x * y else a)
        Do -> (True, a)
        Dont -> (False, a)

main = do
  input <- readInput <$> readFile "input03"
  print $ run (const True) input
  print $ run id input

Sorry for the delay posting this one, Ategon seemed to have it covered, so I forgot :D I will do better.