# --- Day 9: Encoding Error ---
#
# With your neighbor happily enjoying their video game, you turn your attention
# to an open data port on the little screen in the seat in front of you.
#
# Though the port is non-standard, you manage to connect it to your computer
# through the clever use of several paperclips. Upon connection, the port
# outputs a series of numbers (your puzzle input).
#
# The data appears to be encrypted with the eXchange-Masking Addition System
# (XMAS) which, conveniently for you, is an old cypher with an important weakness.
#
# XMAS starts by transmitting a preamble of 25 numbers. After that, each number
# you receive should be the sum of any two of the 25 immediately previous
# numbers. The two numbers will have different values, and there might be more
# than one such pair.
#
# For example, suppose your preamble consists of the numbers 1 through 25 in a
# random order. To be valid, the next number must be the sum of two of those numbers:
#
# 26 would be a valid next number, as it could be 1 plus 25 (or many other pairs, like 2 and 24).
# 49 would be a valid next number, as it is the sum of 24 and 25.
# 100 would not be valid; no two of the previous 25 numbers sum to 100.
# 50 would also not be valid; although 25 appears in the previous 25 numbers, the two numbers in the pair must be different.
#
# Suppose the 26th number is 45, and the first number (no longer an option,
# as it is more than 25 numbers ago) was 20. Now, for the next number to be
# valid, there needs to be some pair of numbers among 1-19, 21-25, or 45 that
# add up to it:
#
# 26 would still be a valid next number, as 1 and 25 are still within the previous 25 numbers.
# 65 would not be valid, as no two of the available numbers sum to it.
# 64 and 66 would both be valid, as they are the result of 19+45 and 21+45 respectively.
#
# Here is a larger example which only considers the previous 5 numbers (and has a preamble of length 5):
#
# 35
# 20
# 15
# 25
# 47
# 40
# 62
# 55
# 65
# 95
# 102
# 117
# 150
# 182
# 127
# 219
# 299
# 277
# 309
# 576
#
# In this example, after the 5-number preamble, almost every number is the sum
# of two of the previous 5 numbers; the only number that does not follow this
# rule is 127.
#
# The first step of attacking the weakness in the XMAS data is to find the
# first number in the list (after the preamble) which is not the sum of two of
# the 25 numbers before it. What is the first number that does not have this
# property?

library(tidyverse)

test_input <- c(35, 20, 15, 25, 47, 40, 62, 55, 65, 95, 102, 117, 150, 182, 127, 219, 299, 277, 309, 576)

# input <- test_input
input <- as.numeric(read_lines("../input/09-01.txt"))

# preamble_length <- 5
preamble_length <- 25
window <- 1:preamble_length
input_start <- preamble_length + 1
no_sum_found <- TRUE

while(no_sum_found && (length(input)>=(preamble_length + 1))) {

  as.data.frame(t(combn(window, 2))) %>%
    mutate(
      val1 = input[V1],
      val2 = input[V2],
      sum = input[V1] + input[V2]
    ) -> pair_sums

  no_sum_found <- any(pair_sums$sum == input[input_start])

  if (no_sum_found) input <- tail(input, -1)

}

(invalid <- input[input_start])


# --- Part Two ---
#
# The final step in breaking the XMAS encryption relies on the invalid number
#  you just found: you must find a contiguous set of at least two numbers in
#  your list which sum to the invalid number from step 1.
#
# Again consider the above example:
#
# 35
# 20
# 15
# 25
# 47
# 40
# 62
# 55
# 65
# 95
# 102
# 117
# 150
# 182
# 127
# 219
# 299
# 277
# 309
# 576
#
# In this list, adding up all of the numbers from 15 through 40 produces the
# invalid number from step 1, 127. (Of course, the contiguous set of numbers
# in your actual list might be much longer.)
#
# To find the encryption weakness, add together the smallest and largest
# number in this contiguous range; in this example, these are 15 and 47,
# producing 62.
#
# What is the encryption weakness in your XMAS-encrypted list of numbers?
# s
# 09-02 -------------------------------------------------------------------

# input <- test_input
input <- as.numeric(read_lines("../input/09-01.txt"))

for (i in seq_along(input)) {

  csum <- cumsum(input[i:length(input)])
  if (any(csum == invalid)) break

}

sum(range(input[i:(i+which(csum == invalid)-1)]))