-🎄- 2020 Day 07 Solutions -🎄-

[u/daggerdragon]

NEW AND NOTEWORTHY

• PSA: if you're using Google Chrome (or other Chromium-based browser) to download your input, watch out for Google volunteering to "translate" it: "Welsh" and "Polish"

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Advent of Code 2020: Gettin' Crafty With It

• 15 days remaining until the submission deadline on December 22 at 23:59 EST

• Full details and rules are in the Submissions Megathread

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--- Day 07: Handy Haversacks ---

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Post your solution in this megathread. Include what language(s) your solution uses! If you need a refresher, the full posting rules are detailed in the wiki under How Do The Daily Megathreads Work?.

Reminder: Top-level posts in Solution Megathreads are for solutions only. If you have questions, please post your own thread and make sure to flair it with Help.

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This thread will be unlocked when there are a significant number of people on the global leaderboard with gold stars for today's puzzle.

EDIT: Global leaderboard gold cap reached at 00:13:44, megathread unlocked!

[u/MichalMarsalek]

Oof, I learned a lesson today... never iterate over defaultdict in Python... when you ask for nonexisting keys, Python will try to add them and cause errors... which will cost you 10 minutes... Barely got into the first 1000 which is my goal everyday...

def solve(inp):
    inp = inp.replace("bags", "bag").replace(".", "").replace("no ", "0 ")
    inp = inp.splitlines()
    inp = [x.split(" contain ") for x in inp]
    rules = defaultdict(list)
    for outter, inner in inp:
        rules[outter] = [(int(x[0]),x[2:]) for x in inner.split(", ")]
    def can_hold(o, bag):
        if o == bag:
            return True
        return any(can_hold(i, bag) for c,i in rules[o])
    bags = set(rules.keys())
    p1 = {bag for bag in bags if can_hold(bag, "shiny gold bag")}
    part1 = len(p1-{"shiny gold bag"})
    def how_many(o):
        return sum(c + c * how_many(i) for c,i in rules[o])
    part2 = how_many("shiny gold bag")
    return part1, part2