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--- Part Two ---
It's getting pretty expensive to fly these days - not because of ticket prices, but because of the
ridiculous number of bags you need to buy!
Consider again your shiny gold bag and the rules from the above example:
- faded blue bags contain 0 other bags. - dotted black bags contain 0 other bags. - vibrant plum
bags contain 11 other bags: 5 faded blue bags and 6 dotted black bags. - dark olive bags contain 7
other bags: 3 faded blue bags and 4 dotted black bags.
So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags within it) plus 2 vibrant
plum bags (and the 11 bags within [1m[37meach[0m of those): 1 + 1*7 + 2 + 2*11 = [1m[37m32[0m
bags!
Of course, the actual rules have a small chance of going several levels deeper than this example; be
sure to count all of the bags, even if the nesting becomes topologically impractical!
Here's another example:
shiny gold bags contain 2 dark red bags. dark red bags contain 2 dark orange bags. dark orange bags
contain 2 dark yellow bags. dark yellow bags contain 2 dark green bags. dark green bags contain 2
dark blue bags. dark blue bags contain 2 dark violet bags. dark violet bags contain no other bags.
In this example, a single shiny gold bag must contain [1m[37m126[0m other bags.
[1m[37mHow many individual bags are required inside your single shiny gold bag?[0m
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