day02.rb

# --- Day 2: Inventory Management System ---
#
# You stop falling through time, catch your breath, and check the screen on the
# device. "Destination reached. Current Year: 1518. Current Location: North Pole Utility Closet
# 83N10." You made it! Now, to find those anomalies.
#
# Outside the utility closet, you hear footsteps and a voice. "...I'm not sure either. But now
# that so many people have chimneys, maybe he could sneak in that way?" Another voice responds,
# "Actually, we've been working on a new kind of suit that would let him fit through tight spaces
# like that. But, I heard that a few days ago, they lost the prototype fabric, the design plans,
# everything! Nobody on the team can even seem to remember important details of the project!"
#
# "Wouldn't they have had enough fabric to fill several boxes in the warehouse? They'd be stored
# together, so the box IDs should be similar. Too bad it would take forever to search the
# warehouse for two similar box IDs..." They walk too far away to hear any more.
#
# Late at night, you sneak to the warehouse - who knows what kinds of paradoxes you could cause if
# you were discovered - and use your fancy wrist device to quickly scan every box and produce a
# list of the likely candidates (your puzzle input).
#
# To make sure you didn't miss any, you scan the likely candidate boxes again, counting the number
# that have an ID containing exactly two of any letter and then separately counting those with
# exactly three of any letter. You can multiply those two counts together to get a rudimentary
# checksum and compare it to what your device predicts.
#
# For example, if you see the following box IDs:
#
#
# abcdef contains no letters that appear exactly two or three times.
# bababc contains two a and three b, so it counts for both.
# abbcde contains two b, but no letter appears exactly three times.
# abcccd contains three c, but no letter appears exactly two times.
# aabcdd contains two a and two d, but it only counts once.
# abcdee contains two e.
# ababab contains three a and three b, but it only counts once.
#
# Of these box IDs, four of them contain a letter which appears exactly twice, and three of them
# contain a letter which appears exactly three times. Multiplying these together produces a
# checksum of 4 * 3 = 12.
#
# What is the checksum for your list of box IDs?
#

require_relative 'input'

day = __FILE__[/\d+/].to_i(10)
input = Input.for_day(day, 2018)

# puts "solving day #{day} from input:\n#{input}"

require_relative '../ext/enumerable'

# --- Part Two ---
#
# Confident that your list of box IDs is complete, you're ready to find the boxes full of
# prototype fabric.
#
# The boxes will have IDs which differ by exactly one character at the same position in both
# strings. For example, given the following box IDs:
#
# abcde
# fghij
# klmno
# pqrst
# fguij
# axcye
# wvxyz
#
# The IDs abcde and axcye are close, but they differ by two characters (the second and
# fourth). However, the IDs fghij and fguij differ by exactly one character, the third (h and
# u). Those must be the correct boxes.
#
# What letters are common between the two correct box IDs? (In the example above, this is found by
# removing the differing character from either ID, producing fgij.)

doubles = 0
triples = 0
boxes = input.each_line(chomp: true).to_a
boxes.each do |box|
  chars = box.chars.count_by
  doubles += 1 if chars.values.include? 2
  triples += 1 if chars.values.include? 3
end
puts "Part 1: #{doubles * triples}"

require 'set'

def diffs(aid, bid)
  aid.chars.zip(bid.chars).partition {|a,b| a == b}.last.size
end

candidates = Set.new
boxes.each do |one|
  if boxes.select {|box| diffs(one, box) == 1}.count == 1
    candidates << one
  end
end

opts = candidates.to_a
match = opts.first.chars.zip(opts.last.chars).flat_map {|a,b| a == b ? a : nil}.compact.join
puts "match: #{match}"