# --- Day 6: Lanternfish ---

# The sea floor is getting steeper. Maybe the sleigh keys got carried this way?

# A massive school of glowing lanternfish swims past. They must spawn quickly
# to reach such large numbers - maybe exponentially quickly? You should model
# their growth rate to be sure.

# Although you know nothing about this specific species of lanternfish, you
# make some guesses about their attributes. Surely, each lanternfish creates a
# new lanternfish once every 7 days.

# However, this process isn't necessarily synchronized between every
# lanternfish - one lanternfish might have 2 days left until it creates another
# lanternfish, while another might have 4. So, you can model each fish as a
# single number that represents the number of days until it creates a new
# lanternfish.

# Furthermore, you reason, a new lanternfish would surely need slightly longer
# before it's capable of producing more lanternfish: two more days for its
# first cycle.

# So, suppose you have a lanternfish with an internal timer value of 3:

#     After one day, its internal timer would become 2.
#     After another day, its internal timer would become 1.
#     After another day, its internal timer would become 0.
#     After another day, its internal timer would reset to 6, and it would
# create a new lanternfish with an internal timer of 8.
#     After another day, the first lanternfish would have an internal timer of
# 5, and the second lanternfish would have an internal timer of 7.

# A lanternfish that creates a new fish resets its timer to 6, not 7 (because 0
# is included as a valid timer value). The new lanternfish starts with an
# internal timer of 8 and does not start counting down until the next day.

# Realizing what you're trying to do, the submarine automatically produces a
# list of the ages of several hundred nearby lanternfish (your puzzle input).
# For example, suppose you were given the following list:

# 3,4,3,1,2

# This list means that the first fish has an internal timer of 3, the second
# fish has an internal timer of 4, and so on until the fifth fish, which has an
# internal timer of 2. Simulating these fish over several days would proceed as
# follows:

# Initial state: 3,4,3,1,2
# After  1 day:  2,3,2,0,1
# After  2 days: 1,2,1,6,0,8
# After  3 days: 0,1,0,5,6,7,8
# After  4 days: 6,0,6,4,5,6,7,8,8
# After  5 days: 5,6,5,3,4,5,6,7,7,8
# After  6 days: 4,5,4,2,3,4,5,6,6,7
# After  7 days: 3,4,3,1,2,3,4,5,5,6
# After  8 days: 2,3,2,0,1,2,3,4,4,5
# After  9 days: 1,2,1,6,0,1,2,3,3,4,8
# After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
# After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
# After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
# After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
# After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
# After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
# After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
# After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
# After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8

# Each day, a 0 becomes a 6 and adds a new 8 to the end of the list, while each
# other number decreases by 1 if it was present at the start of the day.

# In this example, after 18 days, there are a total of 26 fish. After 80 days,
# there would be a total of 5934.

# Find a way to simulate lanternfish. How many lanternfish would there be after
# 80 days?

from collections import defaultdict
from copy import copy
from typing import DefaultDict

with open("files/P6.txt", "r") as f:
    data = [int(number) for number in f.read().strip().split(",")]


def part_1() -> None:
    fishes = data.copy()
    day = 0
    while day < 80:
        next_gen = []
        for fish in fishes:
            if fish == 0:
                # reset to 6
                next_gen.append(fish + 6)
                # create offspring with init state of 8
                next_gen.append(fish + 8)
            else:
                # decrease timer by 1
                next_gen.append(fish - 1)

        fishes = next_gen
        day += 1

    print(
        f"After 80 days, there will be {sum(1 for fish in fishes)} lanterfishes"
    )


# --- Part Two ---

# Suppose the lanternfish live forever and have unlimited food and space. Would
# they take over the entire ocean?

# After 256 days in the example above, there would be a total of 26984457539
# lanternfish!

# How many lanternfish would there be after 256 days?


def part_2() -> None:
    # same algorithm but more efficient data structure
    fishes = data.copy()
    fishes_map: dict[int, int] = defaultdict(int)
    for fish in fishes:
        fishes_map[fish] += 1

    day = 0
    while day < 256:
        next_gen: dict[int, int] = defaultdict(int)
        for k, v in fishes_map.items():
            if k == 0:
                next_gen[k + 6] += v
                next_gen[k + 8] += v
            else:
                next_gen[k - 1] += v

        fishes_map = next_gen
        day += 1

    print(
        f"After 256 days, there will be {sum(fishes_map.values())} lanterfishes"
    )


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