idk how to do multi line spoilers lol
letters = ['','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z', 'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
total_score = 0
team = []
for line in open("input.txt"):
team.append(line.strip())
for i in range(2, len(team), 3):
for letter in team[i]:
if letter in team[i-2] and letter in team[i-1]:
total_score += letters.index(letter)
break
print(total_score)
some real caveman shit but it works
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It's not usually this bad, this is just me being r-slurred
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yeah man, maybe u should have just used a lookup array? tho, idk wtf ur doing really, so yeah.
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For every 3 lines:
one by one split them in half, put each half's characters in a set, find the set intersection i.e. the character that's present in both halves and get its value
take all 3 lines and find the character that's present in all 3 also using sets and get its value
the values get summed up in their respective variables, sum and sum2
It's a bit difficult if you don't know functional programming.
Also cleaned it up a bit:
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fuk man, i wrote my solution in about 10 lines using js split and reduce, with one hashmap. if that's not "functional", then idc for whatever functional is.
but what ur doing is ... u don't need to put the 2nd half in another hashmap just so u can use builtins to find the intersection between the two, lol. u can just iterate over the 2nd half and as soon as u find a character in the hashmap of the 1st half .... that's ur "intersetion". i suppose the built-ins do that for u, but it doesn't look easier to me, and ur creating excess data structures.
but (2) is even more mind boggling. once u have the set intersection, that should be one character, and why do u ever need to make sure it's present in all 3?
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Well yeah, I could've but at this point it's a difference of a couple of lines at most so w/e.
Because that's the point of the puzzle?
In their example 'r' was the solution but if I did what you say, I'd also get 'f'
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if only the problem was a couple extra lines.
and the insection itself guarantees it's in both sets, already. are r u doing additional searches to find something already known?
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I'm not doing any additional searches, after the first intersection, the resulting set only has a few characters. That set is then intersected with the third one. It's not like I'm doing a triple for loop.
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the 1st and 2nd half need to be put into sets and intersected, and that intersection should be a set with 1 character. where is this 2nd intersection even coming from?
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ok, i'm a dumbass and just had no idea about the second part. lol. still have no idea what ur code is doing. like what r backpacks?
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2 intersections are for part 2 of the problem. Part 1 is 1 intersection obviously. I don't get what you're confused by.
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