Day 1: Historian Hysteria

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FAQ

  • lwhjp@lemmy.sdf.org
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    28 days 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
    
  • mykl@lemmy.world
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    28 days 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@programming.dev
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    28 days 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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    28 days 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

  • Rin@lemm.ee
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    28 days ago

    TypeScript

    Solution
    import { AdventOfCodeSolutionFunction } from "./solutions";
    
    function InstancesOf(sorted_array: Array<number>, value: number) {
        const index = sorted_array.indexOf(value);
        if(index == -1)
            return 0;
    
        let sum = 1;
    
        for (let array_index = index + 1; array_index < sorted_array.length; array_index++) {
            if(sorted_array[array_index] != value)
                break;
    
            sum += 1;
        }
    
        return sum;
    }
    
    export const solution_1: AdventOfCodeSolutionFunction = (input) => {
        const left: Array<number> = [];
        const right: Array<number> = [];
    
        const lines = input.split("\n");
    
        for (let index = 0; index < lines.length; index++) {
            const element = lines[index].trim();
            if(!element)
                continue;
    
            const leftRight = element.split("   ");
            left.push(Number(leftRight[0]));
            right.push(Number(leftRight[1]));
        }
    
        const numSort = (a: number, b: number) => a - b;
        left.sort(numSort);
        right.sort(numSort);
    
        let sum = 0;
        for (let index = 0; index < left.length; index++) {
            const leftValue = left[index];
            const rightValue = right[index];
    
            sum += Math.abs(leftValue - rightValue);
        }
    
        const part1 = `Part 1: ${sum}`;
    
        sum = 0;
        for (let index = 0; index < left.length; index++) {
            sum += left[index] * InstancesOf(right, left[index]);
        }
    
        const part2 = `Part 2: ${sum}`;
    
        return `${part1}\n${part2}`;
    };
    

    Not the most elegant solution but it works. Decided to reuse the array since it is sorted for both sides.

  • MalachiAzrael@sh.itjust.works
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    28 days ago

    Python

    Part 1
    left_list = []
    right_list = []
    
    for line in file:
        split_line = line.split()
        left_list.append(int(split_line[0]))
        right_list.append(int(split_line[1]))
    
    sorted_left = sorted(left_list)
    sorted_right = sorted(right_list)
    distance = []
    for left, right in zip(sorted_left, sorted_right):
        distance.append(abs(left - right))
    
    total = sum(distance)
    
    print(total)
    
    Part 2
    file = open('input.txt', 'r')
    left_list = []
    right_list = []
    
    for line in file:
        split_line = line.split()
        left_list.append(int(split_line[0]))
        right_list.append(int(split_line[1]))
    
    sim_score = 0
    for item in left_list:
        sim = right_list.count(item)
        sim_score += (sim * item)
    
    print(sim_score)
    

    I am sure there were better ways to do this, this was just the first way in my head, in the order it appeared

  • sjmulder@lemmy.sdf.org
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    28 days 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@programming.dev
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      28 days 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?

      • sjmulder@lemmy.sdf.org
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        28 days ago

        I’m not sure if I understand the question correctly but for every number in the left array I count in the right array. That means duplicate work but shrug 😅

  • proved_unglue@programming.dev
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    26 days ago

    Kotlin

    No 💜 for Kotlin here?

    import kotlin.math.abs
    
    fun part1(input: String): Int {
        val diffs: MutableList<Int> = mutableListOf()
        val pair = parse(input)
        pair.first.sort()
        pair.second.sort()
        pair.first.forEachIndexed { idx, num ->
            diffs.add(abs(num - pair.second[idx]))
        }
        return diffs.sum()
    }
    
    fun part2(input: String): Int {
        val pair = parse(input)
        val frequencies = pair.second.groupingBy { it }.eachCount()
        var score = 0
        pair.first.forEach { num ->
            score += num * frequencies.getOrDefault(num, 0)
        }
        return score
    }
    
    private fun parse(input: String): Pair<MutableList<Int>, MutableList<Int>> {
        val left: MutableList<Int> = mutableListOf()
        val right: MutableList<Int> = mutableListOf()
        input.lines().forEach { line ->
            if (line.isNotBlank()) {
                val parts = line.split("\\s+".toRegex())
                left.add(parts[0].toInt())
                right.add(parts[1].toInt())
            }
        }
        return left to right
    }
    
    • the_beber@lemm.ee
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      26 days ago

      I have another Kotlin (albeit similar) solution:

      import kotlin.math.abs
      
      fun main() {
      
          fun getLists(input: List<String>): Pair<List<Int>, List<Int>> {
              val unsortedPairs = input.map {
                  it.split("   ").map { it.toInt() }
              }
      
              val listA = unsortedPairs.map { it.first() }
              val listB = unsortedPairs.map { it.last() }
              return Pair(listA, listB)
          }
      
          fun part1(input: List<String>): Int {
              val (listA, listB) = getLists(input)
      
              return listA.sorted().zip(listB.sorted()).sumOf { abs(it.first - it.second) }
          }
      
          fun part2(input: List<String>): Int {
              val (listA, listB) = getLists(input)
      
              return listA.sumOf { number ->
                  number * listB.count { it == number }
              }
          }
      
          // Or read a large test input from the `src/Day01_test.txt` file:
          val testInput = readInput("Day01_test")
          check(part1(testInput) == 11)
          check(part2(testInput) == 31)
      
          // Read the input from the `src/Day01.txt` file.
          val input = readInput("Day01")
          part1(input).println()
          part2(input).println()
      }
      
      

      It’s a bit more compact. (If you take out the part that actually calls the functions on the (test-)input.)

      • proved_unglue@programming.dev
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        25 days ago

        Thanks! I like the Pair destruction and zip().sumOf() approach. I’m relatively new to Kotlin, so this is a good learning experience. 😅

  • Ananace@lemmy.ananace.dev
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    28 days ago

    Not going to push hard on these first days (fever being a reason), so I slept in quite a bit before looking at the problem.

    C#
    List<int> _LeftList = new List<int>();
    List<int> _RightList = new List<int>();
    
    // Fed via File.ReadLines(...).Select(l => l.Trim())
    public void Input(IEnumerable<string> lines)
    {
      foreach (var line in lines)
      {
        var split = line.Split(' ', StringSplitOptions.RemoveEmptyEntries).Select(s => int.Parse(s));
        _LeftList.Add(split.First());
        _RightList.Add(split.Last());
      }
    }
    
    public void Part1()
    {
      Console.WriteLine($"Sum: {_LeftList.Order().Zip(_RightList.Order()).Select(v => Math.Abs(v.First - v.Second)).Sum()}");
    }
    public void Part2()
    {
      Console.WriteLine($"Sum: {_LeftList.Select(l => _RightList.Where(i => i == l).Count() * l).Sum()}");
    }
    
  • Leavingoldhabits@lemmy.world
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    28 days ago

    I’m quite inexperienced as a programmer, I learned most of the basic concepts from playing human resource machine and 7 billion humans. After mucking about writing some CLI utilities in Perl and python, I’ve decided to give rust a go.

    Part 1

    Part 2

  • morrowind@lemmy.ml
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    27 days ago

    Smalltalk

    day1p12: input    
    	| list1 list2 nums dist sim |
    	
    	list1 := OrderedCollection new.
    	list2 := OrderedCollection new.
    	
    	input linesDo: [ :l |
    		nums := l substrings collect: [ :n | n asInteger ].
    		list1 add: (nums at: 1).
    		list2 add: (nums at: 2).
    	].
    
    	list1 sort.
    	list2 sort.
    	
    	dist := 0.
    	list1 with: list2 do: [ :a :b | dist := dist + (a - b) abs ].
    	
    	sim := list1 sumNumbers: [ :x | x * (list2 occurrencesOf: x) ].
    	
    	^ Array with: dist with: sim.
    
  • hades@lemm.ee
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    28 days 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();
    }
    
  • bugsmith@programming.dev
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    28 days ago

    I’m late to the party, as usual. Damned timezones. This year I’m going to tackle with a small handful of languages, but primarily Elixir and Gleam. This is my first time trying this languages in earnest, so expect some terrible, inefficient and totally unidiomatic code!
    Here’s day one:

    Elixir

    part_one =
      File.read!("input.in")
      |> String.split("\n", trim: true)
      |> Enum.map(fn line ->
        line
        |> String.split()
        |> Enum.map(&String.to_integer/1)
      end)
      |> Enum.reduce({[], []}, fn [first, second], {list1, list2} ->
        {[first | list1], [second | list2]}
      end)
      |> then(fn {list1, list2} ->
        {Enum.sort(list1), Enum.sort(list2)}
      end)
      |> then(fn {list1, list2} ->
        Enum.zip(list1, list2)
        |> Enum.map(fn {x, y} -> abs(x - y) end)
      end)
      |> Enum.sum()
    
    part_two =
      File.read!("input.in")
      |> String.split("\n", trim: true)
      |> Enum.map(fn line ->
        line
        |> String.split()
        |> Enum.map(&String.to_integer/1)
      end)
      |> Enum.reduce({[], []}, fn [first, second], {list1, list2} ->
        {[first | list1], [second | list2]}
      end)
      |> then(fn {list1, list2} ->
        Enum.map(list1, fn line ->
          line * Enum.count(list2, fn x -> x === line end)
        end)
        |> Enum.sum()
      end)
    
    IO.inspect(part_one)
    IO.inspect(part_two)
    
  • Euro@programming.dev
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    28 days ago

    Viml

    I think viml is a very fun language, i like weird languages lol, so this year im doing it in viml while trying to use as many of the original ed/ex commands as i can (:d, :p, :a, :g, …)

    Part 1
    !cp ./puzzle1 ./puzzle1.editing
    e ./puzzle1.editing
    
    1,$sort
    let row1 = []
    
    g/^\d/let row1 = add(row1, str2nr(expand("<cword>"))) | norm 0dw
    1d
    1,$sort
    g/^\d/execute 'norm cc' .. string(abs(expand("<cword>") - row1[line('.') - 1]))
    
    $a|---ANSWER---
    0
    .
    1,$-1g/^\d/call setline("$", str2nr(getline("$")) + str2nr(expand("<cword>")))
    
    Part 2
    read ./puzzle1
    
    let cnt = 0
    g/^\d/let cnt += expand("<cword>") *
                \ searchcount(#{pattern: '\s\+' .. expand("<cword>")}).total
    
    echo cnt .. "\n"
    
    w! ./puzzle1.editing
    
  • vole@lemmy.world
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    28 days ago

    Raku

    I’m trying warm up to Raku again.

    Solution

    github

    use v6;
    
    sub MAIN($input) {
        my $file = open $input;
    
        grammar LocationList {
            token TOP { <row>+%"\n" "\n"* }
            token row { <left=.id> " "+ <right=.id> }
            token id { \d+ }
        }
    
        my $locations = LocationList.parse($file.slurp);
        my @rows = $locations<row>.map({ (.<left>.Int, .<right>.Int)});
        my $part-one-solution = (@rows[*;0].sort Z- @rows[*;1].sort)».abs.sum;
        say "part 1: $part-one-solution";
    
        my $rbag = bag(@rows[*;1].sort);
        my $part-two-solution = @rows[*;0].map({ $_ * $rbag{$_}}).sum;
        say "part 2: $part-two-solution";
    }
    

    I’m happy to see that Lemmy no longer eats Raku code.