Here is my, very late, solution for day 6. Part two for this took a lot longer than I had estimated, but that's why it's called an estimate.

The brute force method stops working for large iterations. We don't really need to keep track of each fish and it's internal timer for birth. We can instead aggregate the number of fishes for each internal time count in a dictionary. Keeping track of the aggregate rather than the individuals saves us a lot of compute power, thus allowing us to solve part 2. The nice thing about this day's problem is that both part's need the same logic, the second part requires an optimized solution.

import copy
from collections import Counter

if __name__ == "__main__":
    with open("input.txt", "r") as f:
        fishes = f.read().split("\n")[0].split(",")
    
    fishes = list(map(int, fishes))
    results_for_each_fish = {}
    current_fishes = Counter(fishes)

    default_days_till_new_fish = 6

    for i in range(256):
        updated_fishes = copy.deepcopy(current_fishes)

        new_fishes = 0
        for j in range(9):
            if j == 0:
                new_fishes = updated_fishes.get(0, 0)
                updated_fishes[0] = 0
                continue
            updated_fishes[j - 1] = current_fishes.get(j, 0)

        updated_fishes[default_days_till_new_fish + 2] = new_fishes
        updated_fishes[default_days_till_new_fish] = updated_fishes.get(default_days_till_new_fish, 0) + new_fishes
        current_fishes = copy.deepcopy(updated_fishes)

        if i == 79:
            part_one_fish_count = sum([v for _, v in current_fishes.items()])

    part_two_fish_count = sum([v for _, v in current_fishes.items()])

    print(f"Result for Part 1: {part_one_fish_count}")
    print(f"Result for Part 1: {part_two_fish_count}")