Links
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