Advent of Code 2024 Solutions

All solutions on this page were written by me without reading any hints or solutions.

My solutions aren’t necessarily great, but they did the job and got me the answers.

My suggestion to run these solutions:

$ cat input.txt | ./solution.py

Links to the full challenge specifications are included throughout this document. The original specifications aren’t allowed to be reproduced, so I can’t repost them here.

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Day 1 [Spec]

Part 1 Solution

#!/usr/bin/env python3

from sys import stdin

pairs = [tuple(map(int, s.split())) for s in stdin.readlines()]

(a, b) = (sorted([x[0] for x in pairs]), sorted([x[1] for x in pairs]))
print(sum(abs(x - y) for (x, y) in zip(a, b)))

Part 2 Solution

#!/usr/bin/env python3

from sys import stdin
from collections import Counter

pairs = [tuple(map(int, s.split())) for s in stdin.readlines()]

cnt = Counter(x[1] for x in pairs)
print(sum(x[0] * cnt[x[0]] for x in pairs))

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Day 2 [Spec]

Part 1 Solution

#!/usr/bin/env python3

from sys import stdin
from itertools import pairwise

reports = [[int(x) for x in s.split()] for s in stdin.readlines()]
reports = [r for r in reports if (r == sorted(r) or r == sorted(r, reverse=True))]
print(sum(all(0 < abs(a - b) < 4 for a, b in pairwise(r)) for r in reports))

Part 2 Solution

My first solution:

#!/usr/bin/env python3

from sys import stdin
from itertools import pairwise

solution = 0
for s in stdin.readlines():
    report = [int(x) for x in s.split()]
    for i in range(len(report) + 1):
        r = report[:i] + report[i+1:]
        if (r == sorted(r) or r == sorted(r, reverse=True)) \
                and all(0 < abs(a - b) < 4 for a, b in pairwise(r)):
            solution += 1
            break
print(solution)

In an attempt to make a more compact solution, I reimplemented with a lambda:

#!/usr/bin/env python3

from sys import stdin
from itertools import pairwise

is_safe = lambda r: (r == sorted(r) or r == sorted(r, reverse=True)) \
    and all(0 < abs(a - b) < 4 for a, b in pairwise(r))

reports = [[int(x) for x in s.split()] for s in stdin.readlines()]
print(sum(any(is_safe(r[:i] + r[i+1:]) for i in range(len(r) + 1)) for r in reports))

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Day 3 [Spec]

Part 1 Solution

I cheated by using eval() rather than parsing out the mul() myself:

#!/usr/bin/env python3

from sys import stdin
from re import findall
from operator import mul

lines = "".join(stdin.readlines())
print(sum(eval(s) for s in findall(r"mul\([0-9]{1,3},[0-9]{1,3}\)", lines)))

Part 2 Solution

My first solution, retaining the use of eval():

#!/usr/bin/env python3

from sys import stdin
from re import findall
from operator import mul

lines = "".join(stdin.readlines())
matches = findall(r"(mul\([0-9]{1,3},[0-9]{1,3}\))|(do\(\))|(don't\(\))", lines)

solution = 0
do_mul = True
for a, b, c in matches:
    if a and do_mul:
        solution += eval(a)
    elif b:
        do_mul = True
    elif c:
        do_mul = False
print(solution)

My attempt at making a more compact solution:

#!/usr/bin/env python3

from sys import stdin
from re import findall
from operator import mul

lines = "".join(stdin.readlines())

solution = 0
do_mul = True
for s, do, _ in findall(r"(mul\([0-9]{1,3},[0-9]{1,3}\))|(do\(\))|(don't\(\))", lines):
    if not s:
        do_mul = do
    elif do_mul:
        solution += eval(a)
print(solution)