Advent of Code - Day 2

NobbZ · December 2, 2018, 12:27pm

Note by the Moderators: This topic is to talk about the Day 2 of the Advent of Code.

For general discussion about the Advent of Code 2018 and links to topics of the other days, see this topic.

This is my todays code:

NobbZ/aoc_ex/blob/master/lib/y2018/d2.ex

defmodule AocEx.Y2018.D2 do
  use AocEx.Boilerplate,
    transform: fn raw -> raw |> String.split("\n") end

  def part1(input \\ processed()) do
    {two, three} =
      Enum.reduce(input, {0, 0}, fn id, {two, three} ->
        {c2, c3} =
          id
          |> String.to_charlist()
          |> Enum.group_by(& &1)
          |> Enum.reduce({false, false}, fn
            {_, [_, _]}, {_, three} -> {true, three}
            {_, [_, _, _]}, {two, _} -> {two, true}
            _, acc -> acc
          end)

        {if(c2, do: two + 1, else: two), if(c3, do: three + 1, else: three)}
      end)

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I’m not quite happy with it though, but it runs in ~4ms/~8ms, so I leave it as is for now.

sorentwo · December 2, 2018, 4:35pm

I’m doing my problems in Elixir this year as well: http://github.com/sorentwo/aoc_2018

My solution today seems a little different from the others I’ve seen. I made use of String.myers_difference/2 for the second part:

  def common(addresses) do
    pairings = for add_a <- addresses, add_b <- addresses, add_a != add_b, do: {add_a, add_b}

    Enum.reduce_while(pairings, nil, fn {add_a, add_b}, _ ->
      case String.myers_difference(add_a, add_b) do
        [eq: _, del: << char :: size(8) >>, ins: _, eq: _] ->
          {:halt, String.replace(add_a, IO.chardata_to_string([char]), "")}

        _ -> {:cont, nil}
      end
    end)
  end

There are two other possible edit scripts that could have matched (first character or last character being the difference), but this worked for my inputs.

yawaramin · December 3, 2018, 3:34am

I’m pretty happy with my solutions so far, especially for Day 2 Part 2 I felt I was able to take advantage of a few things, especially a comprehension to structure it nicely:

github.com

yawaramin/advent_of_code_2018/blob/864628064c79d5b6248d098c85b787d7627a41da/lib/day2.ex#L11


defmodule AdventOfCode2018.Day2 do
  def part1 do
    {box_id_repeats_2, box_id_repeats_3} =
      input()
      |> Stream.map(&letter_repeats/1)
      |> Enum.reduce({0, 0}, &sum_repeat_counts/2)


    IO.puts(box_id_repeats_2 * box_id_repeats_3)
  end


  def part2 do
    boxids = Stream.map(input(), &String.trim/1)


    for boxid_1 <- boxids,
        boxid_2 <- boxids,
        boxid_1 < boxid_2,
        diff = String.myers_difference(boxid_1, boxid_2),
        one_char_diff?(diff) do
      diff |> Keyword.get_values(:eq) |> IO.puts()
    end
  end

debajit · December 3, 2018, 6:00am

I had a lot of fun doing Day 2 Part 2.

github.com

debajit/advent-of-code-2018/blob/master/day-02-part-2-inventory.exs

defmodule InventoryManagementSystem do
  @moduledoc """
  Solution to Advent of Code puzzle for Day 2 (Part 2)
  https://adventofcode.com/2018/day/2

  The solution essentially boils down to this:
  - Given a list of words, we need to
  - Compare each pair of words together (O(n^2) “loop”)
  - Find the first pair that differ by one character (Enum.find_value operation in Elixir)
  - Return the common letters
  """

  def find_common_substring(ids) do
    for i <- ids, j <- ids, i != j do
      {i, j}
    end
    |> Enum.find_value(fn {x, y} ->
      find_common_letters_if_strings_differ_by_one_char(x, y)
    end)
  end

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I like how this one turned out — reasonably small and clean.

I’ve learned that finding the right functional operation (Enum.find_value in this case) really helps keep the code small (and easy to reason about). (String.myers_difference helped, of course. I wrote my first version with MapSet (and set difference and intersections), which interestingly worked for the example case, but did not work with the large input file, so I ended up using String.myers_difference)