Day 2: Red-Nosed Reports

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FAQ

1 point

Elixir

defmodule AdventOfCode.Solution.Year2024.Day02 do
  use AdventOfCode.Solution.SharedParse

  @impl true
  def parse(input) do
    for line <- String.split(input, "\n", trim: true),
        do: String.split(line) |> Enum.map(&String.to_integer/1)
  end

  def part1(input) do
    Enum.count(input, &is_safe(&1, false))
  end

  def part2(input) do
    Enum.count(input, &(is_safe(&1, true) or is_safe(tl(&1), false)))
  end

  def is_safe([a, b, c | rest], can_fix) do
    cond do
      (b - a) * (c - b) > 0 and abs(b - a) in 1..3 and abs(c - b) in 1..3 ->
        is_safe([b, c | rest], can_fix)

      can_fix ->
        is_safe([a, c | rest], false) or is_safe([a, b | rest], false)

      true ->
        false
    end
  end

  def is_safe(_, _), do: true
end
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3 points

J

There is probably a way to write this more point-free. You can definitely see here the friction involved in the way J wants to regard lists as arrays: short rows of the input matrix are zero padded, so you have to snip off the padding before you process each row, and that means you can’t lift some of the operations back up to the parent matrix because it will re-introduce the padding as it reshapes the result; this accounts for a lot of the "1 everywhere (you can interpret v"1 as “force the verb v to operate on rank 1 subarrays of the argument”).

data_file_name =: '2.data'
data =: > 0 ". each cutopen toJ fread data_file_name

NB. {. take, i. index of; this removes trailing zeros
remove_padding =: {.~ i.&amp;0

NB. }. behead, }: curtail; this computes successive differences
diff =: }. - }:

NB. a b in_range y == a &lt;: y &lt;: b
in_range =: 4 : '(((0 { x) &amp; &lt;:) * (&lt;: &amp; (1 { x))) y'

NB. a row is safe if either all successive differences are in [1..3] or all in [_3.._1]
NB. +. or
ranges =: 2 2 $ 1 3 _3 _1
row_safe =: (+./"1) @: (*/"1) @: (ranges &amp; (in_range"1 _)) @: diff @: remove_padding

result1 =: +/ safe"1 data

NB. x delete y is y without the xth element
delete =: 4 : '(x {. y) , ((>: x) }. y)'"0 _
modified_row =: 3 : 'y , (i.#y) delete y'

modified_row_safe =: 3 : '+./"1 row_safe"1 modified_row"1 y'
result2 =: +/ modified_row_safe data
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3 points

Rust

use crate::utils::read_lines;

pub fn solution1() {
    let reports = get_reports();
    let safe_reports = reports
        .filter(|report| report.windows(3).all(window_is_valid))
        .count();

    println!("Number of safe reports = {safe_reports}");
}

pub fn solution2() {
    let reports = get_reports();
    let safe_reports = reports
        .filter(|report| {
            (0..report.len()).any(|i| {
                [&report[0..i], &report[i + 1..]]
                    .concat()
                    .windows(3)
                    .all(window_is_valid)
            })
        })
        .count();

    println!("Number of safe reports = {safe_reports}");
}

fn window_is_valid(window: &[usize]) -> bool {
    matches!(window[0].abs_diff(window[1]), 1..=3)
        && matches!(window[1].abs_diff(window[2]), 1..=3)
        && ((window[0] > window[1] && window[1] > window[2])
            || (window[0] < window[1] && window[1] < window[2]))
}

fn get_reports() -> impl Iterator<Item = Vec<usize>> {
    read_lines("src/day2/input.txt").map(|line| {
        line.split_ascii_whitespace()
            .map(|level| {
                level
                    .parse()
                    .expect("Reactor level is always valid integer")
            })
            .collect()
    })
}

Definitely trickier than yesterday’s. I feel like the windows solution isn’t the best, but it was what came to mind and ended up working for me.

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3 points

TypeScript

Solution
import { AdventOfCodeSolutionFunction } from "./solutions";


/**
 * this function evaluates the 
 * @param levels a list to check
 * @returns -1 if there is no errors, or the index of where there's an unsafe event
 */
export function EvaluateLineSafe(levels: Array<number>) {
    // this loop is the checking every number in the line
    let isIncreasing: boolean | null = null;
    for (let levelIndex = 1; levelIndex < levels.length; levelIndex++) {
        const prevLevel = levels[levelIndex - 1]; // previous
        const level = levels[levelIndex]; // current
        const diff = level - prevLevel; // difference
        const absDiff = Math.abs(diff); // absolute difference

        // check if increasing too much or not at all
        if (absDiff == 0 || absDiff > 3)
            return levelIndex; // go to the next report

        // set increasing if needed
        if (isIncreasing === null) {
            isIncreasing = diff > 0;
            continue; // compare the next numbers
        }

        //  check if increasing then decreasing 
        if (!(isIncreasing && diff > 0 || !isIncreasing && diff < 0))
            return levelIndex; // go to the next report
    }

    return -1;
}


export const solution_2: AdventOfCodeSolutionFunction = (input) => {
    const reports = input.split("\n");

    let safe = 0;
    let safe_damp = 0;

    // this loop is for every line
    main: for (let i = 0; i < reports.length; i++) {
        const report = reports[i].trim();
        if (!report)
            continue; // report is empty

        const levels = report.split(" ").map((v) => Number(v));

        const evaluation = EvaluateLineSafe(levels);
        if(evaluation == -1) {
            safe++;
            continue;
        }
        
        // search around where it failed
        for (let offset = evaluation - 2; offset <= evaluation + 2; offset++) {
            // delete an evaluation in accordance to the offset
            let newLevels = [...levels];
            newLevels.splice(offset, 1);
            const newEval = EvaluateLineSafe(newLevels);
            if(newEval == -1) {
                safe_damp++;
                continue main;
            }
        }
    }

    return `Part 1: ${safe} Part 2: ${safe + safe_damp}`;
}

God, I really wish my solutions weren’t so convoluted. Also, this is an O(N^3) solution…

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3 points

I don’t think your solution is O(N^3). Can you explain your reasoning?

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1 point

3 nested for loops

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2 points

It’s not as simple as that. You can have 20 nested for loops with complexity of O(1) if all of them only ever finish one iteration.

Or you can have one for loop that iterates 2^N times.

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2 points

Haskell

Had some fun with arrows.

import Control.Arrow
import Control.Monad

main = getContents >>= print . (part1 &&& part2) . fmap (fmap read . words) . lines

part1 = length . filter isSafe
part2 = length . filter (any isSafe . removeOne)

isSafe = ap (zipWith (-)) tail >>> (all (between 1 3) &&& all (between (-3) (-1))) >>> uncurry (||)
 where
  between a b = (a <=) &&& (<= b) >>> uncurry (&&)

removeOne [] = []
removeOne (x : xs) = xs : fmap (x :) (removeOne xs)
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1 point

I like the branched pipelines in isSafe! Very cute.

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