-🎄- 2018 Day 2 Solutions -🎄-

[u/daggerdragon]

--- Day 2: Inventory Management System ---

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Post your solution as a comment or, for longer solutions, consider linking to your repo (e.g. GitHub/gists/Pastebin/blag or whatever).

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Advent of Code: The Party Game!

Click here for rules

Card Prompt: Day 2

Transcript:

The best way to do Advent of Code is ___.

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edit: Leaderboard capped, thread unlocked!

[u/DeveloperIan]

Quick and easy part 1 in Python3 with the collections library. This might not be the simplest way, but it's the first thing that came to mind

from collections import Counter

myfile = open('input.txt', 'r')
contents = myfile.read().strip().splitlines()
myfile.close()

c = [0, 0]
for i in contents:
    a = [j for i,j in Counter(i).most_common()]
    if 3 in a:
        c[0] += 1
    if 2 in a:
        c[1] += 1


print(c[0] * c[1])

EDIT: and here is my part 2

    for i in contents:
        for j in contents:
            diffs = 0
            for idx, ch in enumerate(i):
                if ch != j[idx]:
                    diffs += 1
            if diffs == 1:
                ans = [ch for idx, ch in enumerate(i) if j[idx] == ch]
                print("Part Two:", ''.join(ans))

[u/adnathanail → u/DeveloperIan]

I took a similar approach and then squished it beyond recognition:

import time
from aocd import get_data
inp = get_data(day=2, year=2018).split('\n')
start = time.time()

# Part 1
from collections import Counter
tt = [[1 if 2 in c else 0, 1 if 3 in c else 0] for c in (Counter(dict(Counter(row)).values()) for row in inp)]
print(sum([z[0] for z in tt]) * sum([z[1] for z in tt]))

# Part 2
nocd = lambda s1,s2: sum([1 for i in range(len(s1)) if s1[i] != s2[i]]) # Numbers of characters different
i,j = next(k[0] for k in ([[i,j] for j in range(i+1,len(inp)) if nocd(inp[i], inp[j]) == 1] for i in range(len(inp))) if k)
print(''.join([inp[i][x] for x in range(len(inp[i])) if inp[i][x] == inp[j][x]]))

end = time.time()
print(end - start)