import sys
from itertools import combinations

# --- 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?


with open("files/P9.txt", "r") as f:
    numbers = [int(num) for num in f.read().strip().split("\n")]


def part_1() -> int:
    window = 25
    for i in range(window, len(numbers)):
        if all(
            [
                x + y != numbers[i]
                for x, y in combinations(numbers[i - window : i], 2)
            ]
        ):
            break

    print(f"The first number without the property is {numbers[i]}")
    return numbers[i]


# --- 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?


def part_2() -> None:
    target = invalid_number
    for i in range(len(numbers)):
        for j in range(i + 1, len(numbers)):
            if sum(numbers[i:j]) == target:
                print(
                    f"The encryption weakness is {min(numbers[i:j]) + max(numbers[i:j])}"
                )
                return


if __name__ == "__main__":
    invalid_number = part_1()
    part_2()