Day 1: Historian Hysteria

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FAQ

7 points
*

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
βŠœβˆ˜βŠΈβ‰ @\n     # Partition at \n.
⊜(β†βˆ΅β‹•)βŠΈβ‰ @\s # Partition at space, parse ints, sort.

&p/+/(⌡-). # Part1 : Get abs differences, sum, print.

&p/+Γ—βŸœ(/+β‰β‰‘βŒ•)Β°βŠ‚ # Part 2 : Count instances, mul out, sum, print.
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7 points

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
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4 points
*

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

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4 points

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!();
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4 points
*

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.

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