A straightforward day 1 for a Javascript / NodeJS solution.

let position = 50;
let answer1 = 0;
let answer2 = 0;

const sequence = require('fs').readFileSync('input-day1.txt', 'utf-8');

sequence.split('\n').forEach(instruction => {
    const turn = instruction.charAt(0);
    let distance = parseInt(instruction.slice(1), 10);

    while (distance > 0) {
        distance -= 1;

        if (turn === 'L') {
            position -= 1;
        } else if (turn === 'R') {
            position += 1;
        }

        if (position >= 100) {
            position -= 100;
        } else if (position < 0) {
            position += 100;
        }

        if (position === 0) {
            answer2 += 1;
        }
    }

    if (position === 0) {
        answer1 += 1;
    }
});

console.log(`Part 1 Answer: ${answer1}`);
console.log(`Part 2 Answer: ${answer2}`);

So, obviously the bot didnt work, thanks Ategon for covering :D

Here is the dumb solution for pt2, everyone can share in my shame:

#[test]
    fn test_2025_1_part2() {
        let input = std::fs::read_to_string("input/2025/day_1.txt").unwrap();
        let turns = input
            .lines()
            .map(|line| {
                let value = line[1..].parse::<i32>().unwrap();
                if line[0..1] == *"L" {
                    -value
                } else {
                    value
                }
            })
            .collect::<Vec<i32>>();
        let mut zero_ctr = 0;
        let mut pos = 50;
        for turn in turns {
            print!("{pos}->");

            let mut zero_passes = 0;
            if turn > 0 {
                for _ in 0..turn {
                    pos += 1;
                    if pos == 100 {
                        pos = 0;
                        zero_passes += 1;
                    }
                }
            } else {
                for _ in 0..-turn {
                    pos -= 1;
                    if pos == 0 {
                        zero_passes += 1;
                    }
                    if pos == -1 {
                        pos = 99;
                    }
                }
            };
            println!("{turn}->{pos}: {zero_passes}");
            zero_ctr += zero_passes;
        }
        println!("zero ctr: {}", zero_ctr);
    }

Rust

#[derive(Default)]
pub struct Day1Solver {
    input: Vec<i64>,
}

impl Solver for Day1Solver {
    fn presolve(&mut self, input: &str) {
        self.input = input
            .trim()
            .split("\n")
            .map(|line| {
                if let Some(n) = line.strip_prefix('L') {
                    -n.parse::<i64>().unwrap()
                } else if let Some(n) = line.strip_prefix('R') {
                    n.parse().unwrap()
                } else {
                    panic!("what: {line}");
                }
            })
            .collect();
    }

    fn solve_part_one(&mut self) -> String {
        let mut p = 50;
        let mut count = 0;
        for n in self.input.clone() {
            p += n;
            if p % 100 == 0 {
                count += 1;
            }
        }
        count.to_string()
    }

    fn solve_part_two(&mut self) -> String {
        let mut count = 0;
        let mut p = 1000000000050;
        for i in self.input.clone() {
            if p % 100 == 0 {
                count += (i / 100).abs();
            } else {
                count += ((p + i) / 100 - p / 100).abs();
                if i < 0 && (p + i) % 100 == 0 {
                    count += 1;
                }
            }
            p += i;
        }
        count.to_string()
    }
}

Rust

Almost missed, that “the dial starts by pointing at 50”.

View on github

const N: i32 = 100;

fn parse_line(l: &str) -> (i32, i32) {
    let dir = match l.chars().next().unwrap() {
        'L' => -1,
        'R' => 1,
        _ => panic!(),
    };
    let dist = l[1..].parse::<i32>().unwrap();
    (dir, dist)
}

fn part1(input: String) {
    let mut pos = 50;
    let mut count0 = 0;
    for l in input.lines() {
        let (dir, dist) = parse_line(l);
        pos = (pos + dir * dist) % N;
        if pos == 0 {
            count0 += 1;
        }
    }
    println!("{count0}");
}

fn part2(input: String) {
    let mut pos = 50;
    let mut count0 = 0;
    for l in input.lines() {
        let (dir, dist) = parse_line(l);
        if dir == 1 {
            count0 += (pos + dist) / N;
        } else {
            count0 += ((N - pos) % N + dist) / N;
        }
        pos = (pos + dir * dist).rem_euclid(N);
    }
    println!("{count0}");
}

util::aoc_main!();

The struggled with a counting solution for a long time. I submitted with a simple enumerative solution in the end but managed to get it right after some pause time:

Haskell

{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE OrPatterns #-}
module Main (main) where

import Control.Monad ( (<$!>) )
import qualified Data.List as List

main :: IO ()
main = do
  rotations <- (fmap parseRotation . init . lines) <$!> getContents
  print $ part1 rotations
  print $ part2 rotations

part2 :: [Either Int Int] -> Int
part2 rotations = let

    foldRotation (position, zeroCount) operation = case operation of
      Left y -> let
        (zeroPasses, y') = y `divMod` 100
        position' = (position - y') `mod` 100
        zeroCount' = zeroPasses + zeroCount + if position <= y' then fromEnum $ position /= 0 else 0
        in (position', zeroCount')
      Right y -> let
        (zeroPasses, y') = y `divMod` 100
        position' = (position + y') `mod` 100
        zeroCount' = zeroPasses + zeroCount + if y' + position >= 100 then 1 else 0
        in (position', zeroCount')

  in snd $ List.foldl' foldRotation (50, 0) rotations

part1 :: [Either Int Int] -> Int
part1 rotations = let
    positions = List.scanl applyRotation 50 rotations
  in List.length . filter (== 0) $ positions

applyRotation :: Int -> Either Int Int -> Int
applyRotation x = \case
  Left y -> (x - y) `mod` 100
  Right y -> (x + y) `mod` 100

parseRotation :: String -> Either Int Int
parseRotation = \case
  'R':rest -> Right $ read rest
  'L':rest -> Left $ read rest
  bad -> error $ "invalid rotation operation: " ++ bad

-- | Old solution enumerating all the numbers

part2' :: [Either Int Int] -> Int
part2' rotations = let
  intermediatePositions _ [] = []
  intermediatePositions x (op:ops) = case op of
    Left 0; Right 0 -> intermediatePositions x ops
    Left y -> let x' = pred x `mod` 100 in x' : intermediatePositions x' (Left (pred y) : ops)
    Right y -> let x' = succ x `mod` 100 in x' : intermediatePositions x' (Right (pred y) : ops)
  in List.length . List.filter (== 0) . intermediatePositions 50 $ rotations

Uiua

Today’s lesson: Never think that scanning the first 100 lines of input will give you a good understanding of it.

Part 2 is really messy and could probably be much simpler, but I couldn’t get the logic straight in my head otherwise.

"L68 L30 R48 L5 R60 L55 L1 L99 R14 L82"
⊜⋕⊸≠@\s∧⍜⊡⋅@¯⊚⊸⌕"L"∧⍜⊡⋅@+⊚⊸⌕"R"
P₁ ← ⧻⊚=0\(◿100+)⊂50
P ← (
  ⊃(-×100×|/+)⌊÷100⌵⟜(⊸±) # Count and remove all over-rotations
  ⍉⊟↘¯1⊸(\(◿100+)⊂50)     # Take all positions and next moves.
  ▽⊸≡(≠0⊢)                # Ignore any starting from zero
  +/+↥⊃(<0|>100)≡/+       # Sum the pairs, check for passing zero.
)
P₂ ← +⊃P P₁
⊃(P₁|P₂)

I agree with strlcpy – computing was better in the 1980s. Let’s try the version of the 1980s where Lisp Machines were going to power the AI future. It can’t be any worse than the AI future we’ve got right now.

(This is SBCL, not a Lisp Machine emulator, because I’m not that hardcore.)

(defun parse-line (line)
  (let ((sign (if (eql (char line 0) #\R) 1 -1))
        (number (parse-integer (subseq line 1))))
    (* sign number)))

(defun read-inputs (filename)
  (let ((input-lines (uiop:read-file-lines filename)))
    (mapcar #'parse-line input-lines)))

(defun rotate (pos rotation)
  (mod (+ pos rotation) 100))

(defun main-1 (filename)
  (let ((rotations (read-inputs filename))
        (pos 50))
    (loop for rotation in rotations
          do (setf pos (rotate pos rotation))
          sum (if (= pos 0) 1 0))))

(defun zero-crossings (pos rotation)
  (if (> rotation 0)
      (floor (+ rotation pos) 100)
      (let ((neg-pos (if (zerop pos) pos (- pos 100))))
        (- (ceiling (+ rotation neg-pos) 100)))))

(defun main-2 (filename)
  (let ((rotations (read-inputs filename))
        (pos 50))
    (loop for rotation in rotations
          sum (zero-crossings pos rotation) into crossings
          do (setf pos (rotate pos rotation))
          finally (return crossings))))

Kotlin

already rather scuffed for day1, but works:

https://codeberg.org/Eco-Gaming/AdventOfCode/src/branch/main/src/main/kotlin/a2025/puzzles/Day01.kt

Nim

That was the rough first day for me. Part 1 was ok. For part 2 I didn’t want to go the easy route, so I was trying to find simple formulaic solution, but my answer was always off by some amount. And debugging was hard, because I I was getting the right answer for example input.
After 40 minutes I wiped everything clean and wrote a bruteforce.

Later that day I returned and solved this one properly. I had to draw many schemes and consider all the edge cases carefully to come up with code below.

type
  AOCSolution[T,U] = tuple[part1: T, part2: U]

proc solve(input: string): AOCSolution[int, int] =
  var dial = 50
  for line in input.splitLines():
    let value = parseInt(line[1..^1])
    let sign = if line[0] == 'L': -1 else: 1
    let offset = value mod 100
    result.part2 += value div 100

    if dial != 0:
      if sign < 0 and offset >= dial or
         sign > 0 and offset >= (100-dial): inc result.part2

    dial = (dial + offset * sign).euclmod(100)
    if dial == 0: inc result.part1

Full solution at Codeberg: solution.nim

Golang

func part1() {
	// file, _ := os.Open("sample.txt")
	file, _ := os.Open("input.txt")
	defer file.Close()

	scanner := bufio.NewScanner(file)
	n := 100
	current := 50
	pointingAt0 := 0

	for scanner.Scan() {
		line := scanner.Text()
		num, _ := strconv.Atoi(line[1:])
		if line[0] == 'L' {
			current = ((current-num)%n + n) % n
		} else {
			current = (current + num) % n
		}
		if current == 0 {
			pointingAt0++
		}
	}

	fmt.Println(pointingAt0)
}

func part2() {
	// file, _ := os.Open("sample.txt")
	file, _ := os.Open("input.txt")
	defer file.Close()

	scanner := bufio.NewScanner(file)
	n := 100
	current := 50
	pointingAt0 := 0

	for scanner.Scan() {
		line := scanner.Text()
		num, _ := strconv.Atoi(line[1:])

		rounds := num / n
		pointingAt0 += rounds
		num = num % n
		new := -1

		if line[0] == 'L' {
			new = ((current-num)%n + n) % n
			if current != 0 && (new > current || new == 0) {
				pointingAt0++
			}
		} else {
			new = (current + num) % n
			if current != 0 && (new < current || new == 0) {
				pointingAt0++
			}
		}

		current = new
	}

	fmt.Println(pointingAt0)
}

Janet

• part1
• part2

Rust

use std::{cmp::Ordering, fs, str::FromStr};

use color_eyre::eyre::{Result, bail};

const DIAL_NUMBERS: isize = 100;

#[derive(Clone, Copy)]
struct Dial(usize);

impl Dial {
    fn new() -> Self {
        Self(50)
    }

    fn rotate(&mut self, rot: isize) -> usize {
        let pass_0s;

        let total = rot.wrapping_add_unsigned(self.0);
        match total.cmp(&0) {
            Ordering::Equal => {
                pass_0s = 1;
                self.0 = 0;
            }
            Ordering::Less => {
                // Starting on 0 means we don't cross it
                let started_0 = if self.0 == 0 { 1 } else { 0 };
                pass_0s = 1 + (-total / DIAL_NUMBERS) as usize - started_0;
                self.0 = (self.0 as isize + rot).rem_euclid(DIAL_NUMBERS) as usize;
            }
            Ordering::Greater => {
                let full_turns = total / DIAL_NUMBERS;
                pass_0s = full_turns as usize;
                self.0 = (total - DIAL_NUMBERS * full_turns) as usize;
            }
        };

        pass_0s
    }

    fn sequence(&mut self, s: &str) -> Result<(usize, usize)> {
        let mut end_0s = 0;
        let mut pass_0s = 0;

        for l in s.lines() {
            let num = isize::from_str(&l[1..]).unwrap();
            let dir = l.bytes().next().unwrap();
            pass_0s += match dir {
                b'L' => self.rotate(-num),
                b'R' => self.rotate(num),
                _ => bail!("b{dir} is an invalid rotation direction"),
            };
            if self.0 == 0 {
                end_0s += 1;
            }
        }

        Ok((end_0s, pass_0s))
    }
}

fn parts(filepath: &str) -> Result<(usize, usize)> {
    let input = fs::read_to_string(filepath)?;
    let mut dial = Dial::new();
    let res = dial.sequence(&input)?;
    Ok(res)
}

fn main() -> Result<()> {
    color_eyre::install()?;

    let (p1, p2) = parts("d01/input.txt")?;
    println!("Part 1: {p1}");
    println!("Part 2: {p2}");
    Ok(())
}

I lost a lot of time struggling with edge cases in part two since I thought it was wasteful to run rem_euclid() (i.e. modulus) when I’d already got the information to do it more efficiently. While I got there in the end my final code was a bit ugly. This prettier, “unoptimised” version runs in 1.6ms, so I was being an idiot.

c

#include "aoc.h"
#include <stdio.h>
#include <string.h>

constexpr usize LINE_BUFSZ = (1 << 3);
constexpr i32 START = 50;
constexpr i32 TOP = 100;

static void
solve(Mode mode) {
  FILE* input = fopen("input", "r");
  c8 line[LINE_BUFSZ] = {};
  u32 zs = 0;
  i32 idx = START;
  while (fgets(line, sizeof(line), input)) {
    line[strcspn(line, "\n")] = 0;
    i32 val = 0;
    sscanf(&line[1], "%d", &val);
    if (mode == MODE_ONE) {
      i32 d = line[0] == 'L' ? -val : val;
      idx = (idx + d) % TOP;
      idx += idx < 0 ? TOP : 0;
      idx == 0 ? zs++ : 0;
    } else {
      for (i32 i = 0; i < val; i++) {
        idx = line[0] == 'L' ? idx - 1 : idx + 1;
        idx = idx == -1 ? TOP - 1 : idx;
        idx = idx == TOP ? 0 : idx;
        idx == 0 ? zs++ : 0;
      }
    }
  }
  fclose(input);
  printf("%u\n", zs);
}

i32
main(void) {
  solve(MODE_ONE);
  solve(MODE_TWO);
}

Haskell

import Control.Arrow
import Control.Monad
import Control.Monad.Writer.Strict
import Data.Char
import Data.Functor
import Text.ParserCombinators.ReadP

n = 100
start = 50

parse = fst . last . readP_to_S (endBy rotation (char '\n'))
  where
    rotation = (*) <$> ((char 'L' $> (-1)) <++ (char 'R' $> 1)) <*> (read <$> munch isDigit)

part1 = length . filter (== 0) . fmap (`mod` n) . scanl (+) start

spins :: Int -> Int -> Writer [Int] Int
spins acc x = do
    when (abs x >= n) . tell . pure $ abs x `div` n -- full loops
    let res = acc + (x `rem` n)
        res' = res `mod` n

    when (res /= res') . tell . pure $ 1

    return res'

part2 = sum . execWriter . foldM spins start

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

DOS + BIOS boot (hybrid binary)

Repo | day01.asm | day01.c (prototype) | .COM download

Written in x86-16 assembly. Works as a DOS program but also as disk image for older PCs with BIOS support (or older VMs). Getting this setup to work was tricky, especially since when I started this, I had only basic x86-16 experience! Needless to say I’ve spent much time staring at hex numbers.

The very start of the file determines if it’s running in DOS or as a bootloader, in which case it’ll have to load the remainder of the file from disk and rearrange the memory layout to emulate the DOS situation.

Right now this is using just one segment of memory (64K). Input data is compressed using a custom run-length encoding scheme. The compressed background image is directly decoded into the VGA framebuffer, after which it is overwritten with the decompressed input file. Space is tight!

My main goal is to finish a few days at least with one or more game consoles supported (GameBoy Advance would be cool) and to do some cool palette tricks, like a day/night transition or animated water and stars.