Rust

Part 1 Part 2

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

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
}

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’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 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)

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")]))

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>

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;
}

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.

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!

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

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.⊔)/↥)+,

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)

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;
}

deleted by creator