Python 3

I had some trouble getting Part 2 to work, until I realized that there could be overlap ( blbleightwoqsqs -> 82).

import re

def puzzle_one():
    result_sum = 0
    with open("inputs/day_01", "r", encoding="utf_8") as input_file:
        for line in input_file:
            number_list = [char for char in line if char.isnumeric()]
            number = int(number_list[0] + number_list[-1])
            result_sum += number
    return result_sum

def puzzle_two():
    regex = r"(?=(zero|one|two|three|four|five|six|seven|eight|nine|[0-9]))"
    number_dict = {
        "zero": "0",
        "one": "1",
        "two": "2",
        "three": "3",
        "four": "4",
        "five": "5",
        "six": "6",
        "seven": "7",
        "eight": "8",
        "nine": "9",
    }
    result_sum = 0
    with open("inputs/day_01", "r", encoding="utf_8") as input_file:
        for line in input_file:
            number_list = [
                number_dict[num] if num in number_dict else num
                for num in re.findall(regex, line)
            ]
            number = int(number_list[0] + number_list[-1])
            result_sum += number
    return result_sum

I still have a hard time understanding regex, but I think it's getting there.

Rust

Part 1 Part 2

Fun with iterators! For the second part I went to regex for help.

A new C solution: without lookahead or backtracking! I keep a running tally of how many letters of each digit word were matched so far: https://github.com/sjmulder/aoc/blob/master/2023/c/day01.c

int main(int argc, char **argv)
{
	static const char names[][8] = {"zero", "one", "two", "three",
	    "four", "five", "six", "seven", "eight", "nine"};
	int p1=0, p2=0, i,c;
	int p1_first = -1, p1_last = -1;
	int p2_first = -1, p2_last = -1;
	int nmatched[10] = {0};
	
	while ((c = getchar()) != EOF)
		if (c == '\n') {
			p1 += p1_first*10 + p1_last;
			p2 += p2_first*10 + p2_last;
			p1_first = p1_last = p2_first = p2_last = -1;
			memset(nmatched, 0, sizeof(nmatched));
		} else if (c >= '0' && c <= '9') {
			if (p1_first == -1) p1_first = c-'0';
			if (p2_first == -1) p2_first = c-'0';
			p1_last = p2_last = c-'0';
			memset(nmatched, 0, sizeof(nmatched));
		} else for (i=0; i<10; i++)
			/* advance or reset no. matched digit chars */
			if (c != names[i][nmatched[i]++])
				nmatched[i] = c == names[i][0];
			/* matched to end? */
			else if (!names[i][nmatched[i]]) {
				if (p2_first == -1) p2_first = i;
				p2_last = i;
				nmatched[i] = 0;
			}

	printf("%d %d\n", p1, p2);
	return 0;
}

Java

My take on a modern Java solution (parts 1 & 2).

package thtroyer.day1;

import java.util.*;
import java.util.stream.IntStream;
import java.util.stream.Stream;


public class Day1 {
    record Match(int index, String name, int value) {
    }

    Map numbers = Map.of(
            "one", 1,
            "two", 2,
            "three", 3,
            "four", 4,
            "five", 5,
            "six", 6,
            "seven", 7,
            "eight", 8,
            "nine", 9);

    /**
     * Takes in all lines, returns summed answer
     */
    public int getCalibrationValue(String... lines) {
        return Arrays.stream(lines)
                .map(this::getCalibrationValue)
                .map(Integer::parseInt)
                .reduce(0, Integer::sum);
    }

    /**
     * Takes a single line and returns the value for that line,
     * which is the first and last number (numerical or text).
     */
    protected String getCalibrationValue(String line) {
        var matches = Stream.concat(
                        findAllNumberStrings(line).stream(),
                        findAllNumerics(line).stream()
                ).sorted(Comparator.comparingInt(Match::index))
                .toList();

        return "" + matches.getFirst().value() + matches.getLast().value();
    }

    /**
     * Find all the strings of written numbers (e.g. "one")
     *
     * @return List of Matches
     */
    private List findAllNumberStrings(String line) {
        return IntStream.range(0, line.length())
                .boxed()
                .map(i -> findAMatchAtIndex(line, i))
                .filter(Optional::isPresent)
                .map(Optional::get)
                .sorted(Comparator.comparingInt(Match::index))
                .toList();
    }


    private Optional findAMatchAtIndex(String line, int index) {
        return numbers.entrySet().stream()
                .filter(n -> line.indexOf(n.getKey(), index) == index)
                .map(n -> new Match(index, n.getKey(), n.getValue()))
                .findAny();
    }

    /**
     * Find all the strings of digits (e.g. "1")
     *
     * @return List of Matches
     */
    private List findAllNumerics(String line) {
        return IntStream.range(0, line.length())
                .boxed()
                .filter(i -> Character.isDigit(line.charAt(i)))
                .map(i -> new Match(i, null, Integer.parseInt(line.substring(i, i + 1))))
                .toList();
    }

    public static void main(String[] args) {
        System.out.println(new Day1().getCalibrationValue(args));
    }
}

I feel ok about part 1, and just terrible about part 2.

day01.factor on github (with comments and imports):

: part1 ( -- )
  "vocab:aoc-2023/day01/input.txt" utf8 file-lines
  [
    [ [ digit? ] find nip ]
    [ [ digit? ] find-last nip ] bi
    2array string>number
  ] map-sum .
;

MEMO: digit-words ( -- name-char-assoc )
  [ "123456789" [ dup char>name "-" split1 nip ,, ] each ] H{ } make
;

: first-digit-char ( str -- num-char/f i/f )
  [ digit? ] find swap
;

: last-digit-char ( str -- num-char/f i/f )
  [ digit? ] find-last swap
;

: first-digit-word ( str -- num-char/f )
  [
    digit-words keys [
      2dup subseq-index
      dup [
        [ digit-words at ] dip
        ,,
      ] [ 2drop ] if
    ] each drop                           !
  ] H{ } make
  [ f ] [
    sort-keys first last
  ] if-assoc-empty
;

: last-digit-word ( str -- num-char/f )
  reverse
  [
    digit-words keys [
      reverse
      2dup subseq-index
      dup [
        [ reverse digit-words at ] dip
        ,,
      ] [ 2drop ] if
    ] each drop                           !
  ] H{ } make
  [ f ] [
    sort-keys first last
  ] if-assoc-empty
;

: first-digit ( str -- num-char )
  dup first-digit-char dup [
    pick 2dup swap head nip
    first-digit-word dup [
      [ 2drop ] dip
    ] [ 2drop ] if
    nip
  ] [
    2drop first-digit-word
  ] if
;

: last-digit ( str -- num-char )
  dup last-digit-char dup [
    pick 2dup swap 1 + tail nip
    last-digit-word dup [
      [ 2drop ] dip
    ] [ 2drop ] if
    nip
  ] [
    2drop last-digit-word
  ] if
;

: part2 ( -- )
  "vocab:aoc-2023/day01/input.txt" utf8 file-lines
  [ [ first-digit ] [ last-digit ] bi 2array string>number ] map-sum .
;

Solution in C: https://github.com/sjmulder/aoc/blob/master/2023/c/day01-orig.c

Usually day 1 solutions are super short numeric things, this was a little more verbose. For part 2 I just loop over an array of digit names and use strncmp().

int main(int argc, char **argv)
{
	static const char * const nm[] = {"zero", "one", "two", "three",
	    "four", "five", "six", "seven", "eight", "nine"};
	char buf[64], *s;
	int p1=0,p2=0, p1f,p1l, p2f,p2l, d;
	
	while (fgets(buf, sizeof(buf), stdin)) {
		p1f = p1l = p2f = p2l = -1;

		for (s=buf; *s; s++)
			if (*s >= '0' && *s <= '9') {
				d = *s-'0';
				if (p1f == -1) p1f = d;
				if (p2f == -1) p2f = d;
				p1l = p2l = d;
			} else for (d=0; d<10; d++) {
				if (strncmp(s, nm[d], strlen(nm[d])))
					continue;
				if (p2f == -1) p2f = d;
				p2l = d;
				break;
			}

		p1 += p1f*10 + p1l;
		p2 += p2f*10 + p2l;
	}

	printf("%d %d\n", p1, p2);
	return 0;
}

Part 02 in Rust 🦀 :

use std::{
    collections::HashMap,
    env, fs,
    io::{self, BufRead, BufReader},
};

fn main() -> io::Result<()> {
    let args: Vec = env::args().collect();
    let filename = &args[1];
    let file = fs::File::open(filename)?;
    let reader = BufReader::new(file);

    let number_map = HashMap::from([
        ("one", "1"),
        ("two", "2"),
        ("three", "3"),
        ("four", "4"),
        ("five", "5"),
        ("six", "6"),
        ("seven", "7"),
        ("eight", "8"),
        ("nine", "9"),
    ]);

    let mut total = 0;
    for _line in reader.lines() {
        let digits = get_text_numbers(_line.unwrap(), &number_map);
        if !digits.is_empty() {
            let digit_first = digits.first().unwrap();
            let digit_last = digits.last().unwrap();
            let mut cat = String::new();
            cat.push(*digit_first);
            cat.push(*digit_last);
            let cat: i32 = cat.parse().unwrap();
            total += cat;
        }
    }
    println!("{total}");
    Ok(())
}

fn get_text_numbers(text: String, number_map: &HashMap<&str, &str>) -> Vec {
    let mut digits: Vec = Vec::new();
    if text.is_empty() {
        return digits;
    }
    let mut sample = String::new();
    let chars: Vec = text.chars().collect();
    let mut ptr1: usize = 0;
    let mut ptr2: usize;
    while ptr1 < chars.len() {
        sample.clear();
        ptr2 = ptr1 + 1;
        if chars[ptr1].is_digit(10) {
            digits.push(chars[ptr1]);
            sample.clear();
            ptr1 += 1;
            continue;
        }
        sample.push(chars[ptr1]);
        while ptr2 < chars.len() {
            if chars[ptr2].is_digit(10) {
                sample.clear();
                break;
            }
            sample.push(chars[ptr2]);
            if number_map.contains_key(&sample.as_str()) {
                let str_digit: char = number_map.get(&sample.as_str()).unwrap().parse().unwrap();
                digits.push(str_digit);
                sample.clear();
                break;
            }
            ptr2 += 1;
        }
        ptr1 += 1;
    }

    digits
}

import re
numbers = {
    "one" : 1,
    "two" : 2,
    "three" : 3,
    "four" : 4,
    "five" : 5,
    "six" : 6,
    "seven" : 7,
    "eight" : 8,
    "nine" : 9
    }
for digit in range(10):
    numbers[str(digit)] = digit
pattern = "(%s)" % "|".join(numbers.keys())
   
re1 = re.compile(".*?" + pattern)
re2 = re.compile(".*" + pattern)
total = 0
for line in open("input.txt"):
    m1 = re1.match(line)
    m2 = re2.match(line)
    num = (numbers[m1.group(1)] * 10) + numbers[m2.group(1)]
    total += num
print(total)

There weren't any zeros in the training data I got - the text seems to suggest that "0" is allowed but "zero" isn't.

Solved part one in about thirty seconds. But wow, either my brain is just tired at this hour or I'm lacking in skill, but part two is harder than any other year has been on the first day. Anyway, I managed to solve it, but I absolutely hate it, and will definitely be coming back to try to clean this one up.

https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day01.rs

impl Solver for Day01 {
    fn star_one(&self, input: &str) -> String {
        let mut result = 0;

        for line in input.lines() {
            let line = line
                .chars()
                .filter(|ch| ch.is_ascii_digit())
                .collect::>();
            let first = line.first().unwrap();
            let last = line.last().unwrap();
            let number = format!("{first}{last}").parse::().unwrap();
            result += number;
        }

        result.to_string()
    }

    fn star_two(&self, input: &str) -> String {
        let mut result = 0;

        for line in input.lines() {
            let mut first = None;
            let mut last = None;

            while first == None {
                for index in 0..line.len() {
                    let line_slice = &line[index..];
                    if line_slice.starts_with("one") || line_slice.starts_with("1") {
                        first = Some(1);
                    } else if line_slice.starts_with("two") || line_slice.starts_with("2") {
                        first = Some(2);
                    } else if line_slice.starts_with("three") || line_slice.starts_with("3") {
                        first = Some(3);
                    } else if line_slice.starts_with("four") || line_slice.starts_with("4") {
                        first = Some(4);
                    } else if line_slice.starts_with("five") || line_slice.starts_with("5") {
                        first = Some(5);
                    } else if line_slice.starts_with("six") || line_slice.starts_with("6") {
                        first = Some(6);
                    } else if line_slice.starts_with("seven") || line_slice.starts_with("7") {
                        first = Some(7);
                    } else if line_slice.starts_with("eight") || line_slice.starts_with("8") {
                        first = Some(8);
                    } else if line_slice.starts_with("nine") || line_slice.starts_with("9") {
                        first = Some(9);
                    }

                    if first.is_some() {
                        break;
                    }
                }
            }

            while last == None {
                for index in (0..line.len()).rev() {
                    let line_slice = &line[index..];
                    if line_slice.starts_with("one") || line_slice.starts_with("1") {
                        last = Some(1);
                    } else if line_slice.starts_with("two") || line_slice.starts_with("2") {
                        last = Some(2);
                    } else if line_slice.starts_with("three") || line_slice.starts_with("3") {
                        last = Some(3);
                    } else if line_slice.starts_with("four") || line_slice.starts_with("4") {
                        last = Some(4);
                    } else if line_slice.starts_with("five") || line_slice.starts_with("5") {
                        last = Some(5);
                    } else if line_slice.starts_with("six") || line_slice.starts_with("6") {
                        last = Some(6);
                    } else if line_slice.starts_with("seven") || line_slice.starts_with("7") {
                        last = Some(7);
                    } else if line_slice.starts_with("eight") || line_slice.starts_with("8") {
                        last = Some(8);
                    } else if line_slice.starts_with("nine") || line_slice.starts_with("9") {
                        last = Some(9);
                    }

                    if last.is_some() {
                        break;
                    }
                }
            }

            result += format!("{}{}", first.unwrap(), last.unwrap())
                .parse::()
                .unwrap();
        }

        result.to_string()
    }
}

Dart solution

This has got to be one of the biggest jumps in trickiness in a Day 1 puzzle. In the end I rolled my part 1 answer into the part 2 logic. [Edit: I've golfed it a bit since first posting it]

import 'package:collection/collection.dart';

var ds = '0123456789'.split('');
var wds = 'one two three four five six seven eight nine'.split(' ');

int s2d(String s) => s.length == 1 ? int.parse(s) : wds.indexOf(s) + 1;

int value(String s, List digits) {
  var firsts = {for (var e in digits) s.indexOf(e): e}..remove(-1);
  var lasts = {for (var e in digits) s.lastIndexOf(e): e}..remove(-1);
  return s2d(firsts[firsts.keys.min]) * 10 + s2d(lasts[lasts.keys.max]);
}

part1(List lines) => lines.map((e) => value(e, ds)).sum;

part2(List lines) => lines.map((e) => value(e, ds + wds)).sum;

I'm a bit late to the party. I forgot about this.

Anyways, my (lazy) C solutions: https://git.sr.ht/~aidenisik/aoc23/tree/master/item/day1

I wanted to see if it was possible to do part 1 in a single line of Python:

print(sum([(([int(i) for i in line if i.isdigit()][0]) * 10 + [int(i) for i in line if i.isdigit()][-1]) for line in open("input.txt")]))

Trickier than expected! I ran into an issue with Lua patterns, so I had to revert to a more verbose solution, which I then used in Hare as well.

Lua:

-- SPDX-FileCopyrightText: 2023 Jummit
--
-- SPDX-License-Identifier: GPL-3.0-or-later

local sum = 0
for line in io.open("1.input"):lines() do
  local a, b = line:match("^.-(%d).*(%d).-$")
  if not a then
    a = line:match("%d+")
    b = a
  end
  if a and b then
    sum = sum + tonumber(a..b)
  end
end
print(sum)

local names = {
  ["one"] = 1,
  ["two"] = 2,
  ["three"] = 3,
  ["four"] = 4,
  ["five"] = 5,
  ["six"] = 6,
  ["seven"] = 7,
  ["eight"] = 8,
  ["nine"] = 9,
  ["1"] = 1,
  ["2"] = 2,
  ["3"] = 3,
  ["4"] = 4,
  ["5"] = 5,
  ["6"] = 6,
  ["7"] = 7,
  ["8"] = 8,
  ["9"] = 9,
}
sum = 0
for line in io.open("1.input"):lines() do
  local firstPos = math.huge
  local first
  for name, num in pairs(names) do
    local left = line:find(name)
    if left and left < firstPos then
      firstPos = left
      first = num
    end
  end
  local last
  for i = #line, 1, -1 do
    for name, num in pairs(names) do
      local right = line:find(name, i)
      if right then
        last = num
        goto found
      end
    end
  end
  ::found::
  sum = sum + tonumber(first * 10 + last)
end
print(sum)

Hare:

// SPDX-FileCopyrightText: 2023 Jummit
//
// SPDX-License-Identifier: GPL-3.0-or-later

use fmt;
use types;
use bufio;
use strings;
use io;
use os;

const numbers: [](str, int) = [
	("one", 1),
	("two", 2),
	("three", 3),
	("four", 4),
	("five", 5),
	("six", 6),
	("seven", 7),
	("eight", 8),
	("nine", 9),
	("1", 1),
	("2", 2),
	("3", 3),
	("4", 4),
	("5", 5),
	("6", 6),
	("7", 7),
	("8", 8),
	("9", 9),
];

fn solve(start: size) void = {
	const file = os::open("1.input")!;
	defer io::close(file)!;
	const scan = bufio::newscanner(file, types::SIZE_MAX);
	let sum = 0;
	for (let i = 1u; true; i += 1) {
		const line = match (bufio::scan_line(&scan)!) {
		case io::EOF =>
			break;
		case let line: const str =>
			yield line;
		};
		let first: (void | int) = void;
		let last: (void | int) = void;
		for (let i = 0z; i < len(line); i += 1) :found {
			for (let num = start; num < len(numbers); num += 1) {
				const start = strings::sub(line, i, strings::end);
				if (first is void && strings::hasprefix(start, numbers[num].0)) {
					first = numbers[num].1;
				};
				const end = strings::sub(line, len(line) - 1 - i, strings::end);
				if (last is void && strings::hasprefix(end, numbers[num].0)) {
					last = numbers[num].1;
				};
				if (first is int && last is int) {
					break :found;
				};
			};
		};
		sum += first as int * 10 + last as int;
	};
	fmt::printfln("{}", sum)!;
};

export fn main() void = {
	solve(9);
	solve(0);
};

[Rust] 11157/6740

use std::fs;

const m: [(&str, u32); 10] = [
    ("zero", 0),
    ("one", 1),
    ("two", 2),
    ("three", 3),
    ("four", 4),
    ("five", 5),
    ("six", 6),
    ("seven", 7),
    ("eight", 8),
    ("nine", 9)
];

fn main() {
    let s = fs::read_to_string("data/input.txt").unwrap();

    let mut u = 0;

    for l in s.lines() {
        let mut h = l.chars();
        let mut f = 0;
        let mut a = 0;

        for n in 0..l.len() {
            let u = h.next().unwrap();

            match u.is_numeric() {
                true => {
                    let v = u.to_digit(10).unwrap();
                    if f == 0 {
                        f = v;
                    }
                    a = v;
                },
                _ => {
                    for (t, v) in m {
                        if l[n..].starts_with(t) {
                            if f == 0 {
                                f = v;
                            }
                            a = v;
                        }
                    }
                },
            }
        }

        u += f * 10 + a;
    }

    println!("Sum: {}", u);
}

Link

Uiua solution

I may add solutions in Uiua depending on how easy I find them, so here's today's (also available to run online):

Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
# if needle is longer than haystack, return zeros
SafeFind ← ((⌕|-.;)< ∩⧻ , ,)
FindDigits ← (× +1⇡9 ⊠(□SafeFind∩⊔) : Inp)
"123456789"
⊜□ ≠@\s . "one two three four five six seven eight nine"
∩FindDigits
BuildNum ← (/+∵(/+⊂⊃(×10↙ 1)(↙ 1⇌) ▽≠0.⊔) /↥)
∩BuildNum+,

or stripping away all the fluff:

Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
⊜□ ≠@\s."one two three four five six seven eight nine" "123456789"
∩(×+1⇡9⊠(□(⌕|-.;)<⊙:∩(⧻.⊔)):Inp)
∩(/+∵(/+⊂⊃(×10↙1)(↙1⇌)▽≠0.⊔)/↥)+,

My solutin in Elixir for both part 1 and part 2 is below. It does use regex and with that there are many different ways to accomplish the goal. I'm no regex master so I made it as simple as possible and relied on the language a bit more. I'm sure there are cooler solutions with no regex too, this is just what I settled on:

defmodule AdventOfCode.Day01 do
  def part1(args) do
    number_regex = ~r/([0-9])/

    args
    |> String.split(~r/\n/, trim: true)
    |> Enum.map(&first_and_last_number(&1, number_regex))
    |> Enum.map(&number_list_to_integer/1)
    |> Enum.sum()
  end

  def part2(args) do
    number_regex = ~r/(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))/

    args
    |> String.split(~r/\n/, trim: true)
    |> Enum.map(&first_and_last_number(&1, number_regex))
    |> Enum.map(fn number -> Enum.map(number, &replace_word_with_number/1) end)
    |> Enum.map(&number_list_to_integer/1)
    |> Enum.sum()
  end

  defp first_and_last_number(string, regex) do
    matches = Regex.scan(regex, string)
    [_, first] = List.first(matches)
    [_, last] = List.last(matches)

    [first, last]
  end

  defp number_list_to_integer(list) do
    list
    |> List.to_string()
    |> String.to_integer()
  end

  defp replace_word_with_number(string) do
    numbers = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]

    String.replace(string, numbers, fn x ->
      (Enum.find_index(numbers, &(&1 == x)) + 1)
      |> Integer.to_string()
    end)
  end
end

My solution in rust. I'm sure there's a lot more clever ways to do it but this is what I came up with.

use std::{io::prelude::*, fs::File, path::Path, io };

fn main() 
{
    run_solution(false); 

    println!("\nPress enter to continue");
    let mut buffer = String::new();
    io::stdin().read_line(&mut buffer).unwrap();

    run_solution(true); 
}

fn run_solution(check_for_spelled: bool)
{
    let data = load_data("data/input");

    println!("\nProcessing Data...");

    let mut sum: u64 = 0;
    for line in data.lines()
    {
        // Doesn't seem like the to_ascii_lower call is needed but... just in case
        let first = get_digit(line.to_ascii_lowercase().as_bytes(), false, check_for_spelled);
        let last = get_digit(line.to_ascii_lowercase().as_bytes(), true, check_for_spelled);


        let num = (first * 10) + last;

        // println!("\nLine: {} -- First: {}, Second: {}, Num: {}", line, first, last, num);
        sum += num as u64;
    }

    println!("\nFinal Sum: {}", sum);
}

fn get_digit(line: &[u8], from_back: bool, check_for_spelled: bool) -> u8
{
    let mut range: Vec = (0..line.len()).collect();
    if from_back
    {
        range.reverse();
    }

    for i in range
    {
        if is_num(line[i])
        {
            return (line[i] - 48) as u8;
        }

        if check_for_spelled
        {
            if let Some(num) = is_spelled_num(line, i)
            {
                return num;
            }
        }
    }

    return 0;
}

fn is_num(c: u8) -> bool
{
    c >= 48 && c <= 57
}

fn is_spelled_num(line: &[u8], start: usize) -> Option
{
    let words = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"];

    for word_idx in 0..words.len()
    {
        let mut i = start;
        let mut found = true;
        for c in words[word_idx].as_bytes()
        {
            if i < line.len() && *c != line[i]
            {
                found = false;
                break;
            }
            i += 1;
        }

        if found && i <= line.len()
        {
            return Some(word_idx as u8 + 1);
        }
    }

    return None;
}

fn load_data(file_name: &str) -> String
{
    let mut file = match File::open(Path::new(file_name))
    {
        Ok(file) => file,
        Err(why) => panic!("Could not open file {}: {}", Path::new(file_name).display(), why),
    };

    let mut s = String::new();
    let file_contents = match file.read_to_string(&mut s) 
    {
        Err(why) => panic!("couldn't read {}: {}", Path::new(file_name).display(), why),
        Ok(_) => s,
    };
    
    return file_contents;
}

Ruby

https://github.com/snowe2010/advent-of-code/blob/master/ruby_aoc/2023/day01/day01.rb

Part 1

execute(1, test_file_suffix: "p1") do |lines|
  lines.inject(0) do |acc, line|
    d = line.gsub(/\D/,'')
    acc += (d[0] + d[-1]).to_i
  end
end

Part 2

map = {
  "one": 1,
  "two": 2,
  "three": 3,
  "four": 4,
  "five": 5,
  "six": 6,
  "seven": 7,
  "eight": 8,
  "nine": 9,
}

execute(2) do |lines|
  lines.inject(0) do |acc, line|
    first_num = line.sub(/(one|two|three|four|five|six|seven|eight|nine)/) do |key|
      map[key.to_sym]
    end
    last_num = line.reverse.sub(/(enin|thgie|neves|xis|evif|ruof|eerht|owt|eno)/) do |key|
      map[key.reverse.to_sym]
    end

    d = first_num.chars.select { |num| numeric?(num) }
    e = last_num.chars.select { |num| numeric?(num) }
    acc += (d[0] + e[0]).to_i
  end
end

Then of course I also code golfed it, but didn't try very hard.

P1 Code Golf

execute(1, alternative_text: "Code Golf 60 bytes", test_file_suffix: "p1") do |lines|
  lines.inject(0){|a,l|d=l.gsub(/\D/,'');a+=(d[0]+d[-1]).to_i}
end

P2 Code Golf (ignore the formatting, I just didn't want to reformat to remove all the spaces, and it's easier to read this way.)

execute(1, alternative_text: "Code Golf 271 bytes", test_file_suffix: "p1") do |z|
  z.inject(0) { |a, l|
    w = %w(one two three four five six seven eight nine)
    x = w.join(?|)
    f = l.sub(/(#{x})/) { |k| map[k.to_sym] }
    g = l.reverse.sub(/(#{x.reverse})/) { |k| map[k.reverse.to_sym] }
    d = f.chars.select { |n| n.match?(/\d/) }
    e = g.chars.select { |n| n.match?(/\d/) }
    a += (d[0] + e[0]).to_i
  }
end

Part 1 felt fairly pretty simple in Haskell:

import Data.Char (isDigit)

main = interact solve

solve :: String -> String
solve = show . sum . map (read . (\x -> [head x, last x]) . filter isDigit) . lines

Part 2 was more of a struggle, though I'm pretty happy with how it turned out. I ended up using concatMap inits . tails to generate all substrings, in order of appearance so one3m becomes ["","o","on","one","one3","one3m","","n","ne","ne3","ne3m","","e","e3","e3m","","3","3m","","m",""]. I then wrote a function stringToDigit :: String -> Maybe Char which simultaneously filtered out the digits and standardised them as Chars.

import Data.List (inits, tails)
import Data.Char (isDigit, digitToInt)
import Data.Maybe (mapMaybe)

main = interact solve

solve :: String -> String
solve = show . sum . map (read . (\x -> [head x, last x]) . mapMaybe stringToDigit . concatMap inits . tails) . lines
--                             |string of first&last digit| |find all the digits |   |all substrings of line|

stringToDigit "one"   = Just '1'
stringToDigit "two"   = Just '2'
stringToDigit "three" = Just '3'
stringToDigit "four"  = Just '4'
stringToDigit "five"  = Just '5'
stringToDigit "six"   = Just '6'
stringToDigit "seven" = Just '7'
stringToDigit "eight" = Just '8'
stringToDigit "nine"  = Just '9'
stringToDigit [x]
  | isDigit x         = Just x
  | otherwise         = Nothing
stringToDigit _       = Nothing

I went a bit excessively Haskell with it, but I had my fun!

This is my solution in Nim:

import strutils, strformat

const digitStrings = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]

### Type definiton for a proc to extract a calibration function from a line
type CalibrationExtracter = proc (line:string): int

## extract a calibration value by finding the first and last numerical digit, and concatenating them
proc extractCalibration1(line:string): int =
  var first,last = -1
    
  for i, c in line:
    if c.isDigit:
      last = parseInt($c)
      if first == -1:
        first = last
        
  result = first * 10 + last

## extract a calibration value by finding the first and last numerical digit OR english lowercase word for a digit, and concatenating them
proc extractCalibration2(line:string): int =
  var first,last = -1
  
  for i, c in line:
    if c.isDigit:
      last = parseInt($c)
      if first == -1:
        first = last
    else: #not a digit parse number words
      for dsi, ds in digitStrings:
        if i == line.find(ds, i):
          last = dsi+1
          if first == -1:
            first = last
          break #break out of digit strings
        
  result = first * 10 + last

### general purpose extraction proc, accepts an extraction function for specific line handling
proc extract(fileName:string, extracter:CalibrationExtracter, verbose:bool): int =
  
  let lines = readFile(fileName).strip().splitLines();
  
  for lineIndex, line in lines:
    if line.len == 0:
      continue
    
    let value = extracter(line)
    result += value
    
    if verbose:
      echo &"Extracted {value} from line {lineIndex} {line}"

### public interface for puzzle part 1
proc part1*(input:string, verbose:bool = false): int =
  result = input.extract(extractCalibration1, verbose);

### public interface for puzzle part 2
proc part2*(input:string, verbose:bool = false): int =
  result = input.extract(extractCalibration2, verbose);

Have a snippet of Ruby, something I hacked together as part of a solution during the WFH morning meeting;

class String
  def to_numberstring(digits_only: false)
    tokens = { 
      one: 1, two: 2, three: 3,
      four: 4, five: 5, six: 6,
      seven: 7, eight: 8, nine: 9
    }.freeze
    ret = ""

    i = 0
    loop do
      if self[i] =~ /\d/
        ret += self[i]
      elsif !digits_only
        tok = tokens.find { |k, _| self[i, k.size] == k.to_s }
        ret += tok.last.to_s if tok
      end
      
      i += 1
      break if i >= size
    end

    ret
  end
end

Crystal. Second one was a pain.

part 1

input = File.read("./input.txt").lines

sum = 0
input.each do |line|
	digits = line.chars.select(&.number?)
	next if digits.empty?
	num = "#{digits[0]}#{digits[-1]}".to_i
	sum += num
end
puts sum

part 2

numbers = {
	"one"=> "1",
	"two"=> "2",
	"three"=> "3",
	"four"=> "4",
	"five"=> "5",
	"six"=> "6",
	"seven"=> "7",
	"eight"=> "8",
	"nine"=> "9",
}
input.each do |line|
	start = ""
	last = ""
	line.size.times do |i|
		if line[i].number?
			start = line[i]
			break
		end
		if i < line.size - 2 && line[i..(i+2)] =~ /one|two|six/
			start = numbers[line[i..(i+2)]]
			break
		end

		if i < line.size - 3 && line[i..(i+3)] =~ /four|five|nine/
			start = numbers[line[i..(i+3)]]
			break
		end 
		
		if i < line.size - 4 && line[i..(i+4)] =~ /three|seven|eight/
			start = numbers[line[i..(i+4)]]
			break
		end 
	end
	
	(1..line.size).each do |i|
		if line[-i].number?
			last = line[-i]
			break
		end
		if i > 2 && line[(-i)..(-i+2)] =~ /one|two|six/
			last = numbers[line[(-i)..(-i+2)]]
			break
		end

		if i > 3 && line[(-i)..(-i+3)] =~ /four|five|nine/
			last = numbers[line[(-i)..(-i+3)]]
			break
		end 
		
		if i > 4 && line[(-i)..(-i+4)] =~ /three|seven|eight/
			last = numbers[line[(-i)..(-i+4)]]
			break
		end 
	end
	sum += "#{start}#{last}".to_i
end
puts sum

Damn, lemmy's tabs are massive

Python

Questions and feedback welcome!

import re

from .solver import Solver

class Day01(Solver):
  def __init__(self):
    super().__init__(1)
    self.lines = []

  def presolve(self, input: str):
    self.lines = input.rstrip().split('\n')

  def solve_first_star(self):
    numbers = []
    for line in self.lines:
      digits = [ch for ch in line if ch.isdigit()]
      numbers.append(int(digits[0] + digits[-1]))
    return sum(numbers)

  def solve_second_star(self):
    numbers = []
    digit_map = {
      "one": 1, "two": 2, "three": 3, "four": 4, "five": 5,
      "six": 6, "seven": 7, "eight": 8, "nine": 9, "zero": 0,
      }
    for i in range(10):
      digit_map[str(i)] = i
    for line in self.lines:
      digits = [digit_map[digit] for digit in re.findall(
          "(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))", line)]
      numbers.append(digits[0]*10 + digits[-1])
    return sum(numbers)

Just getting my feet wet with coding after a decade of 0 programming. CS just didn't work out for me in school, so I swapped over to math. Trying to use Python on my desktop, with Notepad++ and Windows Shell.

with open('01A_input.txt', 'r') as file:
    data = file.readlines()
    
print(data)
NumericList=[]

for row in data:
    word=row
    while not(word[0].isnumeric()):
        word=word[1:]
    while not(word[-1].isnumeric()):
        word=word[:-1]
    #print(word)
    tempWord=word[0]+word[-1]
    NumericList.append(int(tempWord))
    #print(NumericList)
Total=sum(NumericList)
print(Total)

with open('01A_input.txt', 'r') as file:
    data = file.readlines()
    
#print(data)
NumericList=[]
NumberWords=("one", "two", "three", "four", "five", "six", "seven", "eight", "nine")

def wordreplaceleft(wrd):
    if wrd.startswith("one"):
        return "1" + wrd[3:]
    elif wrd.startswith("two"):
        return "2" + wrd[3:]
    elif wrd.startswith("three"):
        return "3" + wrd[5:]
    elif wrd.startswith("four"):
        return "4" + wrd[4:]
    elif wrd.startswith("five"):
        return "5" + wrd[4:]
    elif wrd.startswith("six"):
        return "6" + wrd[3:]
    elif wrd.startswith("seven"):
        return "7" + wrd[5:]
    elif wrd.startswith("eight"):
        return "8" + wrd[5:]
    elif wrd.startswith("nine"):
        return "9" + wrd[4:]

def wordreplaceright(wrd):
    if wrd.endswith("one"):
        return wrd[:-3]+"1"
    elif wrd.endswith("two"):
        return wrd[:-3]+"2"
    elif wrd.endswith("three"):
        return wrd[:-5]+"3"
    elif wrd.endswith("four"):
        return wrd[:-4]+"4"
    elif wrd.endswith("five"):
        return wrd[:-4]+"5"
    elif wrd.endswith("six"):
        return wrd[:-3]+"6"
    elif wrd.endswith("seven"):
        return wrd[:-5]+"7"
    elif wrd.endswith("eight"):
        return wrd[:-5]+"8"
    elif wrd.endswith("nine"):
        return wrd[:-4]+"9"

for row in data:
    wordleft=row
    wordright=row
    
    if wordleft.startswith(NumberWords):
        wordleft=wordreplaceleft(wordleft)
    while not(wordleft[0].isnumeric()):
        if wordleft.startswith(NumberWords):
            wordleft=wordreplaceleft(wordleft)
        else:
            wordleft=wordleft[1:]
            
    if wordright.endswith(NumberWords):
        wordright=wordreplaceright(wordright)
    while not(wordright[-1].isnumeric()):
        if wordright.endswith(NumberWords):
            wordright=wordreplaceright(wordright)
        else:
            wordright=wordright[:-1]
    
    # while not(word[-1].isnumeric()):
        # word=word[:-1]
    # print(word)
    tempWord=wordleft[0]+wordright[-1]
    NumericList.append(int(tempWord))
    #print(NumericList)
Total=sum(NumericList)
print(Total)

Python 3

I'm trying to practice writing clear, commented, testable functions, so I added some things that are strictly unnecessary for the challenge (docstrings, error raising, type hints, tests...), but I think it's a necessary exercise for me. If anyone has comments or criticism about my attempt at "best practices," please let me know!

Also, I thought it was odd that the correct answer to part 2 requires that you allow for overlapping letters such as "threeight", but that doesn't occur in the sample input. I imagine that many people will hit a wall wondering why their answer is rejected.

import re
from pathlib import Path


DIGITS = [
    "zero",
    "one",
    "two",
    "three",
    "four",
    "five",
    "six",
    "seven",
    "eight",
    "nine",
    r"\d",
]

PATTERN_PART_1 = r"\d"
PATTERN_PART_2 = f"(?=({'|'.join(DIGITS)}))"


def get_digit(s: str) -> int:
    """Return the digit in the input

    Args:
        s (str): one string containing a single digit represented by a single arabic numeral or spelled out in lower-case English

    Returns:
        int: the digit as an integer value
    """

    try:
        return int(s)
    except ValueError:
        return DIGITS.index(s)


def calibration_value(line: str, pattern: str) -> int:
    """Return the calibration value in the input

    Args:
        line (str): one line containing a calibration value
        pattern (str): the regular expression pattern to match

    Raises:
        ValueError: if no digits are found in the line

    Returns:
        int: the calibration value
    """

    digits = re.findall(pattern, line)

    if digits:
        return get_digit(digits[0]) * 10 + get_digit(digits[-1])

    raise ValueError(f"No digits found in: '{line}'")


def calibration_sum(lines: str, pattern: str) -> int:
    """Return the sum of the calibration values in the input

    Args:
        lines (str): one or more lines containing calibration values

    Returns:
        int: the sum of the calibration values
    """

    sum = 0

    for line in lines.split("\n"):
        sum += calibration_value(line, pattern)

    return sum


if __name__ == "__main__":
    path = Path(__file__).resolve().parent / "input" / "day01.txt"

    lines = path.read_text().strip()

    print("Sum of calibration values:")
    print(f"• Part 1: {calibration_sum(lines, PATTERN_PART_1)}")
    print(f"• Part 2: {calibration_sum(lines, PATTERN_PART_2)}")

import pytest
from advent_2023_python.day01 import (
    calibration_value,
    calibration_sum,
    PATTERN_PART_1,
    PATTERN_PART_2,
)


LINES_PART_1 = [
    ("1abc2", 12),
    ("pqr3stu8vwx", 38),
    ("a1b2c3d4e5f", 15),
    ("treb7uchet", 77),
]
BLOCK_PART_1 = (
    "\n".join([line[0] for line in LINES_PART_1]),
    sum(line[1] for line in LINES_PART_1),
)

LINES_PART_2 = [
    ("two1nine", 29),
    ("eightwothree", 83),
    ("abcone2threexyz", 13),
    ("xtwone3four", 24),
    ("4nineeightseven2", 42),
    ("zoneight234", 14),
    ("7pqrstsixteen", 76),
]
BLOCK_PART_2 = (
    "\n".join([line[0] for line in LINES_PART_2]),
    sum(line[1] for line in LINES_PART_2),
)


def test_part_1():
    for line in LINES_PART_1:
        assert calibration_value(line[0], PATTERN_PART_1) == line[1]

    assert calibration_sum(BLOCK_PART_1[0], PATTERN_PART_1) == BLOCK_PART_1[1]


def test_part_2_with_part_1_values():
    for line in LINES_PART_1:
        assert calibration_value(line[0], PATTERN_PART_2) == line[1]

    assert calibration_sum(BLOCK_PART_1[0], PATTERN_PART_2) == BLOCK_PART_1[1]


def test_part_2_with_part_2_values():
    for line in LINES_PART_2:
        assert calibration_value(line[0], PATTERN_PART_2) == line[1]

    assert calibration_sum(BLOCK_PART_2[0], PATTERN_PART_2) == BLOCK_PART_2[1]


def test_no_digits():
    with pytest.raises(ValueError):
        calibration_value("abc", PATTERN_PART_1)

    with pytest.raises(ValueError):
        calibration_value("abc", PATTERN_PART_2)

I did this in C. First part was fairly trivial, iterate over the line, find first and last number, easy.

Second part had me a bit worried i would need a more string friendly library/language, until i worked out that i can just strstr to find "one", and then in place switch that to "o1e", and so on. Then run part1 code over the modified buffer. I originally did "1ne", but overlaps such as "eightwo" meant that i got the 2, but missed the 8.

#include 
#include 
#include 
#include 
#include 

size_t readfile(char* fname, char* buffer, size_t buffer_len)
{

    int f = open(fname, 'r');
    assert(f >= 0);
    size_t total = 0;
    do {
        size_t nr = read(f, buffer + total, buffer_len - total);
        if (nr == 0) {
            return total;
        }
        total += nr;
    }
    while (buffer_len - total > 0);
    return -1;
}

int part1(const char* buffer, size_t buffer_len)
{
    int first = -1;
    int last = -1;
    int total = 0;
    for (int i = 0; i < buffer_len; i++)
    {
        char c = buffer[i];
        if (c == '\n')
        {
            if (first == -1) {
                continue;
            }
            total += (first*10 + last);
            first = last = -1;
            continue;
        }
        int val = c - '0';
        if (val > 9 || val < 0)
        {
            continue;
        }
        if (first == -1)
        {
            first = last = val;
        }
        else
        {
            last = val;
        }
    }
    return total;
}

void part2_sanitize(char* buffer, size_t len)
{
    char* p = NULL;
    while ((p = strnstr(buffer, "one", len)) != NULL)
    {
        p[1] = '1';
    }
    while ((p = strnstr(buffer, "two", len)) != NULL)
    {
        p[1] = '2';
    }
    while ((p = strnstr(buffer, "three", len)) != NULL)
    {
        p[1] = '3';
    }
    while ((p = strnstr(buffer, "four", len)) != NULL)
    {
        p[1] = '4';
    }
    while ((p = strnstr(buffer, "five", len)) != NULL)
    {
        p[1] = '5';
    }
    while ((p = strnstr(buffer, "six", len)) != NULL)
    {
        p[1] = '6';
    }
    while ((p = strnstr(buffer, "seven", len)) != NULL)
    {
        p[1] = '7';
    }
    while ((p = strnstr(buffer, "eight", len)) != NULL)
    {
        p[1] = '8';
    }
    while ((p = strnstr(buffer, "nine", len)) != NULL)
    {
        p[1] = '9';
    }
    while ((p = strnstr(buffer, "zero", len)) != NULL)
    {
        p[1] = '0';
    }
}

int main(int argc, char** argv)
{
    assert(argc == 2);
    char buffer[1000000];
    size_t len = readfile(argv[1], buffer, sizeof(buffer));
    {
        int total = part1(buffer, len);
        printf("Part 1 total: %i\n", total);
    }

    {
        part2_sanitize(buffer, len);
        int total = part1(buffer, len);
        printf("Part 2 total: %i\n", total);
    }
}

[Language: C#]

This isn't the most performant or elegant, it's the first one that worked. I have 3 kids and a full time job. If I get through any of these, it'll be first pass through and first try that gets the correct answer.

Part 1 was very easy, just iterated the string checking if the char was a digit. Ditto for the last, by reversing the string. Part 2 was also not super hard, I settled on re-using the iterative approach, checking each string lookup value first (on a substring of the current char), and if the current char isn't the start of a word, then checking if the char was a digit. Getting the last number required reversing the string and the lookup map.

Part 1:

var list = new List((await File.ReadAllLinesAsync(@".\Day 1\PuzzleInput.txt")));

int total = 0;
foreach (var item in list)
{
    //forward
    string digit1 = string.Empty;
    string digit2 = string.Empty;


    foreach (var c in item)
    {
        if ((int)c >= 48 && (int)c <= 57)
        {
            digit1 += c;
        
            break;
        }
    }
    //reverse
    foreach (var c in item.Reverse())
    {
        if ((int)c >= 48 && (int)c <= 57)
        {
            digit2 += c;

            break;
        }

    }
    total += Int32.Parse(digit1 +digit2);
}

Console.WriteLine(total);

Part 2:

var list = new List((await File.ReadAllLinesAsync(@".\Day 1\PuzzleInput.txt")));
var numbers = new Dictionary() {
    {"one" ,   1}
    ,{"two" ,  2}
    ,{"three" , 3}
    ,{"four" , 4}
    ,{"five" , 5}
    ,{"six" , 6}
    ,{"seven" , 7}
    ,{"eight" , 8}
    , {"nine" , 9 }
};
int total = 0;
string digit1 = string.Empty;
string digit2 = string.Empty;
foreach (var item in list)
{
    //forward
    digit1 = getDigit(item, numbers);
    digit2 = getDigit(new string(item.Reverse().ToArray()), numbers.ToDictionary(k => new string(k.Key.Reverse().ToArray()), k => k.Value));
    total += Int32.Parse(digit1 + digit2);
}

Console.WriteLine(total);

string getDigit(string item,                 Dictionary numbers)
{
    int index = 0;
    int digit = 0;
    foreach (var c in item)
    {
        var sub = item.AsSpan(index++);
        foreach(var n in numbers)
        {
            if (sub.StartsWith(n.Key))
            {
                digit = n.Value;
                goto end;
            }
        }

        if ((int)c >= 48 && (int)c <= 57)
        {
            digit = ((int)c) - 48;
            break;
        }
    }
    end:
    return digit.ToString();
}

Did this in Odin (very hashed together, especially finding the last number in part 2):

package day1

import "core:fmt"
import "core:strings"
import "core:strconv"
import "core:unicode"

p1 :: proc(input: []string) {
    total := 0

    for line in input {
        firstNum := line[strings.index_proc(line, unicode.is_digit):][:1]
        lastNum := line[strings.last_index_proc(line, unicode.is_digit):][:1]

        calibrationValue := strings.concatenate({firstNum, lastNum})
        defer delete(calibrationValue)

        num, ok := strconv.parse_int(calibrationValue)

        total += num
    }

    // daggonit thought it was the whole numbers
    /*
    for line in input {
        firstNum := line

        fFrom := strings.index_proc(firstNum, unicode.is_digit)
        firstNum = firstNum[fFrom:]

        fTo := strings.index_proc(firstNum, proc(r:rune)->bool {return !unicode.is_digit(r)})
        if fTo == -1 do fTo = len(firstNum)
        firstNum = firstNum[:fTo]


        lastNum := line
        lastNum = lastNum[:strings.last_index_proc(lastNum, unicode.is_digit)+1]
        lastNum = lastNum[strings.last_index_proc(lastNum, proc(r:rune)->bool {return !unicode.is_digit(r)})+1:]

        calibrationValue := strings.concatenate({firstNum, lastNum})
        defer delete(calibrationValue)

        num, ok := strconv.parse_int(calibrationValue, 10)
        if !ok {
            fmt.eprintf("%s could not be parsed from %s", calibrationValue, line)
            return
        }

        total += num;
    }
    */

    fmt.println(total)
}

p2 :: proc(input: []string) {
    parse_wordable :: proc(s: string) -> int {
        if len(s) == 1 {
            num, ok := strconv.parse_int(s)
            return num
        } else do switch s {
            case "one"  : return 1
            case "two"  : return 2
            case "three": return 3
            case "four" : return 4
            case "five" : return 5
            case "six"  : return 6
            case "seven": return 7
            case "eight": return 8
            case "nine" : return 9
        }

        return -1
    }

    total := 0

    for line in input {
        firstNumI, firstNumW := strings.index_multi(line, {
            "one"  , "1",
            "two"  , "2",
            "three", "3",
            "four" , "4",
            "five" , "5",
            "six"  , "6",
            "seven", "7",
            "eight", "8",
            "nine" , "9",
        })
        firstNum := line[firstNumI:][:firstNumW]


        // last_index_multi doesn't seem to exist, doing this as backup
        lastNumI, lastNumW := -1, -1
        for {
            nLastNumI, nLastNumW := strings.index_multi(line[lastNumI+1:], {
                "one"  , "1",
                "two"  , "2",
                "three", "3",
                "four" , "4",
                "five" , "5",
                "six"  , "6",
                "seven", "7",
                "eight", "8",
                "nine" , "9",
            })

            if nLastNumI == -1 do break

            lastNumI += nLastNumI+1
            lastNumW  = nLastNumW
        }
        lastNum := line[lastNumI:][:lastNumW]

        total += parse_wordable(firstNum)*10 + parse_wordable(lastNum)
    }

    fmt.println(total)
}

Had a ton of trouble with part 1 until I realized I misinterpreted it. Especially annoying because the example was working fine. So paradoxically part 2 was easier than 1.

Part-A in Python: https://github.com/pbui/advent-of-code-2023/blob/master/day01/day01-A.py

Was able to use a list comprehension to read the input.

Part-B in Python: https://github.com/pbui/advent-of-code-2023/blob/master/day01/day01-B.py

This was trickier...

Update: Modified Part 2 to be more functional again by using a map before I filter

[Language: Lean4]

I'll only post the actual parsing and solution. I have written some helpers which are in other files, as is the main function. For the full code, please see my github repo.

Part 2 is a bit ugly, but I'm still new to Lean4, and writing it this way (structural recursion) just worked without a proof or termination.

def parse (input : String) : Option $ List String :=
  some $ input.split Char.isWhitespace |> List.filter (not ∘ String.isEmpty)

def part1 (instructions : List String) : Option Nat :=
  let firstLast := λ (o : Option Nat × Option Nat) (c : Char) ↦
    let digit := match c with
    | '0' => some 0
    | '1' => some 1
    | '2' => some 2
    | '3' => some 3
    | '4' => some 4
    | '5' => some 5
    | '6' => some 6
    | '7' => some 7
    | '8' => some 8
    | '9' => some 9
    | _ => none
    if let some digit := digit then
      match o.fst with
      | none => (some digit, some digit)
      | some _ => (o.fst, some digit)
    else
      o
  let scanLine := λ (l : String) ↦ l.foldl firstLast (none, none)
  let numbers := instructions.mapM ((uncurry Option.zip) ∘ scanLine)
  let numbers := numbers.map λ l ↦ l.map λ (a, b) ↦ 10*a + b
  numbers.map (List.foldl (.+.) 0)

def part2 (instructions : List String) : Option Nat :=
  -- once I know how to prove stuff propery, I'm going to improve this. Maybe.
  let instructions := instructions.map String.toList
  let updateState := λ (o : Option Nat × Option Nat) (n : Nat) ↦ match o.fst with
    | none => (some n, some n)
    | some _ => (o.fst, some n)
  let extract_digit := λ (o : Option Nat × Option Nat) (l : List Char) ↦
    match l with
    | '0' :: _ | 'z' :: 'e' :: 'r' :: 'o' :: _ => (updateState o 0)
    | '1' :: _ | 'o' :: 'n' :: 'e' :: _ => (updateState o 1)
    | '2' :: _ | 't' :: 'w' :: 'o' :: _ => (updateState o 2)
    | '3' :: _ | 't' :: 'h' :: 'r' :: 'e' :: 'e' :: _ => (updateState o 3)
    | '4' :: _ | 'f' :: 'o' :: 'u' :: 'r' :: _ => (updateState o 4)
    | '5' :: _ | 'f' :: 'i' :: 'v' :: 'e' :: _ => (updateState o 5)
    | '6' :: _ | 's' :: 'i' :: 'x' :: _ => (updateState o 6)
    | '7' :: _ | 's' :: 'e' :: 'v' :: 'e' :: 'n' :: _ => (updateState o 7)
    | '8' :: _ | 'e' :: 'i' :: 'g' :: 'h' :: 't' :: _ => (updateState o 8)
    | '9' :: _ | 'n' :: 'i' :: 'n' :: 'e' :: _ => (updateState o 9)
    | _ => o
  let rec firstLast := λ (o : Option Nat × Option Nat) (l : List Char) ↦
    match l with
    | [] => o
    | _ :: cs => firstLast (extract_digit o l) cs
  let scanLine := λ (l : List Char) ↦ firstLast (none, none) l
  let numbers := instructions.mapM ((uncurry Option.zip) ∘ scanLine)
  let numbers := numbers.map λ l ↦ l.map λ (a, b) ↦ 10*a + b
  numbers.map (List.foldl (.+.) 0)

My solution is not pretty at all, but it does the job: Golang

I decided to learn zig this year. Usually I solve both parts in the same source file, but it was annoying so here part 1 and part 2