Rust

Part 1 Part 2

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

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.

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;
}

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.

I finally got my solutions done. I used rust. I feel like 114 lines (not including empty lines or driver code) for both solutions is pretty decent. If lemmy’s code blocks are hard to read, I also put my solutions on github.

use std::{
    cell::OnceCell,
    collections::{HashMap, VecDeque},
    ops::ControlFlow::{Break, Continue},
};

use crate::utils::read_lines;

#[derive(Clone, Copy, PartialEq, Eq)]
enum NumType {
    Digit,
    DigitOrWord,
}

#[derive(Clone, Copy, PartialEq, Eq)]
enum FromDirection {
    Left,
    Right,
}

const WORD_NUM_MAP: OnceCell> = OnceCell::new();

fn init_num_map() -> HashMap<&'static str, u8> {
    HashMap::from([
        ("one", b'1'),
        ("two", b'2'),
        ("three", b'3'),
        ("four", b'4'),
        ("five", b'5'),
        ("six", b'6'),
        ("seven", b'7'),
        ("eight", b'8'),
        ("nine", b'9'),
    ])
}

const MAX_WORD_LEN: usize = 5;

fn get_digit<i>(mut bytes: I, num_type: NumType, from_direction: FromDirection) -> Option
where
    I: Iterator,
{
    let digit = bytes.try_fold(VecDeque::new(), |mut byte_queue, byte| {
        if byte.is_ascii_digit() {
            Break(byte)
        } else if num_type == NumType::DigitOrWord {
            if from_direction == FromDirection::Left {
                byte_queue.push_back(byte);
            } else {
                byte_queue.push_front(byte);
            }

            let word = byte_queue
                .iter()
                .map(|&amp;byte| byte as char)
                .collect::();

            for &amp;key in WORD_NUM_MAP
                .get_or_init(init_num_map)
                .keys()
                .filter(|k| k.len() &lt;= byte_queue.len())
            {
                if word.contains(key) {
                    return Break(*WORD_NUM_MAP.get_or_init(init_num_map).get(key).unwrap());
                }
            }

            if byte_queue.len() == MAX_WORD_LEN {
                if from_direction == FromDirection::Left {
                    byte_queue.pop_front();
                } else {
                    byte_queue.pop_back();
                }
            }

            Continue(byte_queue)
        } else {
            Continue(byte_queue)
        }
    });

    if let Break(byte) = digit {
        Some(byte)
    } else {
        None
    }
}

fn process_digits(x: u8, y: u8) -> u16 {
    ((10 * (x - b'0')) + (y - b'0')).into()
}

fn solution(num_type: NumType) {
    if let Ok(lines) = read_lines("src/day_1/input.txt") {
        let sum = lines.fold(0_u16, |acc, line| {
            let line = line.unwrap_or_else(|_| String::new());
            let bytes = line.bytes();
            let left = get_digit(bytes.clone(), num_type, FromDirection::Left).unwrap_or(b'0');
            let right = get_digit(bytes.rev(), num_type, FromDirection::Right).unwrap_or(left);

            acc + process_digits(left, right)
        });

        println!("{sum}");
    }
}

pub fn solution_1() {
    solution(NumType::Digit);
}

pub fn solution_2() {
    solution(NumType::DigitOrWord);
}
```</i>

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
}

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;
}

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);
};

I think I found a decently short solution for part 2 in python:

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


def find_digit(word: str) -> str:
    for digit, value in DIGITS.items():
        if word.startswith(digit):
            return value
        if word.startswith(value):
            return value
    return ''


total = 0
for line in puzzle.split('\n'):
    digits = [
        digit for i in range(len(line))
        if (digit := find_digit(line[i:]))
    ]
    total += int(digits[0] + digits[-1])


print(total)

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()
    }
}

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

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));
    }
}

[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

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;
}

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")]))