Advent Of Code
An unofficial home for the advent of code community on programming.dev!
Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.
AoC 2023
Solution Threads
M | T | W | T | F | S | S |
---|---|---|---|---|---|---|
1 | 2 | 3 | ||||
4 | 5 | 6 | 7 | 8 | 9 | 10 |
11 | 12 | 13 | 14 | 15 | 16 | 17 |
18 | 19 | 20 | 21 | 22 | 23 | 24 |
25 |
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console.log('Hello World')
I had some trouble getting Part 2 to work, until I realized that there could be overlap ( blbleightwoqsqs -> 82).
spoiler
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;
}
And golfed down:
char*N[]={0,"one","two","three","four","five","six","seven","eight","nine"};p,P,
i,c,a,b;A,B;m[10];main(){while((c=getchar())>0){c==10?p+=a*10+b,P+=A*10+B,a=b=A=
B=0:0;c>47&&c<58?b=B=c-48,a||(a=b),A||(A=b):0;for(i=10;--i;)c!=N[i][m[i]++]?m[i]
=c==*N[i]:!N[i][m[i]]?A||(A=i),B=i:0;}printf("%d %d\n",p,P);
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(|&byte| byte as char)
.collect::();
for &key in WORD_NUM_MAP
.get_or_init(init_num_map)
.keys()
.filter(|k| k.len() <= 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>
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 02 in Rust π¦ :
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
}
Thanks, used this as input for reading the Day 2 file and looping the lines, just getting started with rust :)
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:
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:
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'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).
spoiler
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);
}
Oh, doing this is Rust is really simple.
I tried doing the same thing in Rust, but ended up doing it in Python instead.
Started a bit late due to setting up the thread and monitoring the leaderboard to open it up but still got it decently quick for having barely touched rust
Probably able to get it down shorter so might revisit it
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;
}
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;
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()
}
}
there where only two rules about posting here and you managed to break one of them
edit: oh sry, only one rule
Wow, I sure did. I was tired last night. I'll edit my solution in and fix my error.
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)
Looks very elegant! I'm having trouble understanding how this finds digit "words" from the end of the line though, as they should be spelled backwards IIRC?
I.e. eno
, owt
, eerht
It simply finds all possible digits and then locates the last one through the reverse indexing. Itβs not efficient because thereβs no shortcircuiting, but thereβs no need to search the strings backwards, which is nice. Pythonβs startswith method is also hiding a lot of that other implementations have done explicitly.
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.
Code
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;
}
One small thing that would make it easier to read would be to use (I don't know what the syntax is called) as opposed to the magic numbers for getting the ascii code for a character as a u8
, e.g. b'0'
instead of 48
.
Oh yeah that's a good point. I have seen that before but I didn't think of it at the time.
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 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 .
;
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
Thank you for sharing this. I also wrote a regular expression with \d|eno|owt
and so on, and I was not so proud of myself :). Good to know I wasn't the only one :).
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 β {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
# if needle is longer than haystack, return zeros
SafeFind β ((β|-.;)< β©β§» , ,)
FindDigits β (Γ +1β‘9 β (β‘SafeFindβ©β) : Inp)
"123456789"
ββ‘ β @\s . "one two three four five six seven eight nine"
β©FindDigits
BuildNum β (/+β΅(/+ββ(Γ10β 1)(β 1β) β½β 0.β) /β₯)
β©BuildNum+,
or stripping away all the fluff:
Inp β {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"}
ββ‘ β @\s."one two three four five six seven eight nine" "123456789"
β©(Γ+1β‘9β (β‘(β|-.;)<β:β©(⧻.β)):Inp)
β©(/+β΅(/+ββ(Γ10β1)(β1β)β½β 0.β)/β₯)+,
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);
}
}
Just realised how inefficient the sanitize function is, it iterates over the buffer way too many times. Should be restarting the strnstr from the location of the last hit instead of from the start.
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 Char
s.
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!
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.
day01.py
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"β’ Part 1: {calibration_sum(lines, PATTERN_PART_1)}")
print(f"β’ Part 2: {calibration_sum(lines, PATTERN_PART_2)}")
test_day01.py
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)
Did this in Odin (very hashed together, especially finding the last number in part 2):
spoiler
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.
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);
Oh hey, a fellow nim person. Have you joined the community?
Here's mine. Kbin doesn't even support code blocks, so using topaz:
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
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.
Part 1
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)
Part 2
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
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)
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...
Hint
You have to account for overlapping words such as: eightwo
. This actually simplifies things as you just need to go letter by letter and check if it is a digit or one of the words.
Update: Modified Part 2 to be more functional again by using a map
before I filter
Re. your hint, that's funny because I skipped the obvious optimization of skipping over matched letters out of laziness, but it would've actually broken the solution.
[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();
}