Advent of Code Day 3

Links

• Intro
• Problem Statement
• Code

Highlights

• As in most programming languages, strings are represented as binary code points in Elixir. The ? operator can be used to match on a range of characters, and the <<>> can be used to get the code of a string.
• For the first part, I split each line in half and checked to see if each item in the first half was in the second half using a reduce. As soon as a match is found, the value is calculated and replaces the accumulator. Any subsequent calls to reduce simply return the accumulator as is if it is already an integer.
• For the second part, I chunked the lines by three. For each group of three, I got rid of duplicates within each sack and flattened all three into a single list. Then I calculated the frequencies of each character and found the character with a frequency of three.

defmodule Day03 do use AOC def value(code) when code in ?a..?z, do: code - 96 def value(code) when code in ?A..?Z, do: code - 38 def part1 do input(3) ~> String.split("\n") ~> Enum.map(fn sack -> middle = div(String.length(sack), 2) {first, second} = String.split_at(sack, middle) first ~> String.to_charlist() ~> Enum.reduce(false, fn _, acc when is_integer(acc) -> acc char, _ -> String.contains?(second, <<char>>) && value(char) end) end) ~> Enum.sum end def part2 do input(3) ~> String.split("\n") ~> Enum.chunk_every(3) ~> Enum.map(fn chunk -> chunk ~> Enum.flat_map(&Enum.uniq(String.to_charlist(&1))) ~> Enum.frequencies() ~> Enum.find(fn {_, freq} -> freq == 3 end) ~> fn {char, _} -> value(char) end end) ~> Enum.sum end end