import aoclib
import math

DAY = 13
TEST_SOLUTION_PART1 = 295
TEST_SOLUTION_PART2 = 1068781


def part1(test_mode=False):
    my_input = aoclib.getInputAsArraySplit(day=DAY, split_char=",", test=test_mode)

    my_arrival = int(my_input[0][0])
    bus_ids = [int(i) for i in my_input[1] if i != 'x']
    waiting_times = [i - my_arrival % i for i in bus_ids]
    min_waiting_time = min(waiting_times)

    return min_waiting_time * bus_ids[waiting_times.index(min_waiting_time)]


def part2(test_mode=False):
    my_input = aoclib.getInputAsArraySplit(day=DAY, split_char=",", test=test_mode)

    bus_ids = {i: int(v) for i, v in enumerate(my_input[1]) if v != 'x'}
    # utilizing the Chinese Remainder Theorem we are searching for x = i (mod bus_ids[i]) for all i
    # watch https://www.youtube.com/watch?v=ru7mWZJlRQg for an easy explanation

    # finding the "left" part of each (mod x) part
    base_multiplier = {i: math.prod(bus_ids.values()) // v for i, v in bus_ids.items()}

    # finding the "right" part of each (mod x) part utilizing the Extended Euclidean Algorithm
    # s. Python documentation on pow(x, -1, y)
    ext_multiplier = {i: pow(base_multiplier[i], -1, bus_ids[i]) for i in bus_ids}

    # sum all multiplications together and add our offset
    # EEA gives us base_multiplier[i] * x == 1(one!) (mod bus_ids[i])
    # but we need base_multiplier[i] * x == i (mod bus_ids[i])
    answer = sum(i * base_multiplier[i] * ext_multiplier[i] for i in bus_ids)

    # and shrink it down
    # for the -answer see pythons behaviour when calculating the mod of negative numbers with positive divisor
    # also: http://python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html
    lowest_common_multiple = math.lcm(*bus_ids.values())
    answer = -answer % lowest_common_multiple

    return answer