require_relative 'util'

# --- Day 6: Signals and Noise ---

# Something is jamming your communications with Santa. Fortunately,
# your signal is only partially jammed, and protocol in situations
# like this is to switch to a simple repetition code to get the
# message through.

# In this model, the same message is sent repeatedly. You've recorded
# the repeating message signal (your puzzle input), but the data seems
# quite corrupted - almost too badly to recover. Almost.

# All you need to do is figure out which character is most frequent
# for each position. For example, suppose you had recorded the
# following messages:

#     eedadn
#     drvtee
#     eandsr
#     raavrd
#     atevrs
#     tsrnev
#     sdttsa
#     rasrtv
#     nssdts
#     ntnada
#     svetve
#     tesnvt
#     vntsnd
#     vrdear
#     dvrsen
#     enarar

# The most common character in the first column is e; in the second,
# a; in the third, s, and so on. Combining these characters returns
# the error-corrected message, `easter`.

# Given the recording in your puzzle input, what is the
# error-corrected version of the message being sent?

input = File.open('06.txt') { |f| f.readlines.map(&:chomp) }
test_input = "eedadn\ndrvtee\neandsr\nraavrd\natevrs\ntsrnev\nsdttsa\nrasrtv
nssdts\nntnada\nsvetve\ntesnvt\nvntsnd\nvrdear\ndvrsen\nenarar".split("\n")

def most_common(chars)
  histogram = Hash.new { |h, k| h[k] = 0 }
  chars.each { |char| histogram[char] += 1 }
  histogram.max_by { |_, v| v }[0]
end

def easy(input)
  input.map(&:chars).transpose.map { |chars| most_common(chars) }.join
end

assert(easy(test_input) == 'easter')
puts "easy(input): #{easy(input)}"

# --- Part Two ---

# Of course, that would be the message - if you hadn't agreed to use a
# modified repetition code instead.

# In this modified code, the sender instead transmits what looks like
# random data, but for each character, the character they actually
# want to send is slightly less likely than the others. Even after
# signal-jamming noise, you can look at the letter distributions in
# each column and choose the least common letter to reconstruct the
# original message.

# In the above example, the least common character in the first column
# is a; in the second, d, and so on. Repeating this process for the
# remaining characters produces the original message, advent.

# Given the recording in your puzzle input and this new decoding
# methodology, what is the original message that Santa is trying to
# send?

def least_common(chars)
  histogram = Hash.new { |h, k| h[k] = 0 }
  chars.each { |char| histogram[char] += 1 }
  histogram.min_by { |_, v| v }[0]
end

def hard(input)
  input.map(&:chars).transpose.map { |chars| least_common(chars) }.join
end

assert(hard(test_input) == 'advent')
puts "hard(input): #{hard(input)}"