Day 1: Historian Hysteria

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  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
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FAQ

  • mykl@lemmy.world
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    8 minutes ago

    Uiua

    For entertainment purposes only, I’ll be trying a solution in Uiua each day until it all gets too much for me…

    $ 3   4
    $ 4   3
    $ 2   5
    $ 1   3
    $ 3   9
    $ 3   3
    ⊜∘⊸≠@\n     # Partition at \n.
    ⊜(⍆∵⋕)⊸≠@\s # Partition at space, parse ints, sort.
    
    &p/+/(⌵-). # Part1 : Get abs differences, sum, print.
    
    &p/+×⟜(/+⍉≡⌕)°⊂ # Part 2 : Count instances, mul out, sum, print.
    
  • Gobbel2000
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    2 hours ago

    Rust

    Right IDs are directly read into a hash map counter.

    use std::str::FromStr;
    use std::collections::HashMap;
    
    fn part1(input: String) {
        let mut left = Vec::new();
        let mut right = Vec::new();
        for line in input.lines() {
            let mut parts = line.split_whitespace()
                .map(|p| u32::from_str(p).unwrap());
            left.push(parts.next().unwrap());
            right.push(parts.next().unwrap());
        }
        left.sort_unstable();
        right.sort_unstable();
        let diff: u32 = left.iter().zip(right)
            .map(|(l, r)| l.abs_diff(r))
            .sum();
        println!("{diff}");
    }
    
    fn part2(input: String) {
        let mut left = Vec::new();
        let mut right: HashMap<u32, u32> = HashMap::new();
        for line in input.lines() {
            let mut parts = line.split_whitespace()
                .map(|p| u32::from_str(p).unwrap());
            left.push(parts.next().unwrap());
            *right.entry(parts.next().unwrap()).or_default() += 1;
        }
        let similar: u32 = left.iter()
            .map(|n| n * right.get(n).copied().unwrap_or_default())
            .sum();
        println!("{similar}");
    }
    
    util::aoc_main!();
    
  • janAkali@lemmy.one
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    2 hours ago

    Nim

    I’ve got my first sub-1000 rank today (998 for part 2). Yay!
    Simple and straightforward challenge, very fitting for 1st day. Gonna enjoy it while it lasts.

    proc solve(input: string): AOCSolution[int, int] =
      var l1,l2: seq[int]
      for line in input.splitLines():
        let pair = line.splitWhitespace()
        l1.add parseInt(pair[0])
        l2.add parseInt(pair[1])
      l1.sort()
      l2.sort()
    
      block p1:
        for i in 0..l1.high:
          result.part1 += abs(l1[i] - l2[i])
    
      block p2:
        for n in l1:
          result.part2 += n * l2.count(n)
    

    Codeberg repo

  • Leo Uino@lemmy.sdf.org
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    2 hours ago

    Haskell

    Plenty of scope for making part 2 faster, but I think simple is best here. Forgot to sort the lists in the first part, which pushed me waaay off the leaderboard.

    import Data.List
    
    main = do
      [as, bs] <- transpose . map (map read . words) . lines <$> readFile "input01"
      print . sum $ map abs $ zipWith (-) (sort as) (sort bs)
      print . sum $ map (\a -> a * length (filter (== a) bs)) as
    
  • hades@lemm.ee
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    3 hours ago

    C#

    public class Day01 : Solver
    {
      private ImmutableArray<int> left;
      private ImmutableArray<int> right;
    
      public void Presolve(string input)
      {
        var pairs = input.Trim().Split("\n").Select(line => Regex.Split(line, @"\s+"));
        left = pairs.Select(item => int.Parse(item[0])).ToImmutableArray();
        right = pairs.Select(item => int.Parse(item[1])).ToImmutableArray();
      }
    
      public string SolveFirst() => left.Sort().Zip(right.Sort()).Select((pair) => int.Abs(pair.First - pair.Second)).Sum().ToString();
    
      public string SolveSecond() => left.Select((number) => number * right.Where(v => v == number).Count()).Sum().ToString();
    }
    
  • mcmodknower
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    3 hours ago

    Solution in ruby

    This is my third program in ruby after the ruby tutorial and half of the rails tutorial, so don’t expect anything too good from it.

    Also i did this today since i had time, i will probably not comment every day.

  • sjmulder@lemmy.sdf.org
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    4 hours ago

    Solution in C

    Part 1 is a sort and a quick loop. Part 2 could be efficient with a lookup table but it was practically instant with a simple non-memoized scan so left it that way.

    • mcmodknower
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      3 hours ago

      You are using some interesting techniques there. I never imaged you could use the result of == for adding to a counter.

      But how did you handle duplicates in part 2?

  • Sleepless One@lemmy.ml
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    1 hour ago

    Rust

    I’m doing it in Rust again this year. I stopped keeping up with it after day 3 last year, so let’s hope I last longer this time around.

    Solution Spoiler Alert
    use std::collections::HashMap;
    
    use crate::utils::read_lines;
    
    pub fn solution1() {
        let (mut id_list1, mut id_list2) = get_id_lists();
    
        id_list1.sort();
        id_list2.sort();
    
        let total_distance = id_list1
            .into_iter()
            .zip(id_list2)
            .map(|(left, right)| (left - right).abs())
            .sum::<i32>();
    
        println!("Total distance = {total_distance}");
    }
    
    pub fn solution2() {
        let (id_list1, id_list2) = get_id_lists();
    
        let id_count_map = id_list2
            .into_iter()
            .fold(HashMap::<_, i32>::new(), |mut map, id| {
                *map.entry(id).or_default() += 1i32;
    
                map
            });
    
        let similarity_score = id_list1
            .into_iter()
            .map(|id| id * id_count_map.get(&id).copied().unwrap_or_default())
            .sum::<i32>();
    
        println!("Similarity score = {similarity_score}");
    }
    
    fn get_id_lists() -> (Vec<i32>, Vec<i32>) {
        read_lines("src/day1/input.txt")
            .map(|line| {
                let mut ids = line.split_whitespace().map(|id| {
                    id.parse::<i32>()
                        .expect("Ids from input must be valid integers")
                });
    
                (
                    ids.next().expect("First Id on line must be present"),
                    ids.next().expect("Second Id on line must be present"),
                )
            })
            .unzip()
    }
    

    I’m keeping my solutions up on GitHub.