this post was submitted on 04 Dec 2023
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2023

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M T W T F S S
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18 19 20 21 22 23 24
25

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Day 4: Scratchcards


Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ or pastebin (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)

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[–] hades@lemm.ee 3 points 11 months ago

Python

Questions and feedback welcome!

import collections
import re

from .solver import Solver

class Day04(Solver):
  def __init__(self):
    super().__init__(4)
    self.cards = []

  def presolve(self, input: str):
    lines = input.rstrip().split('\n')
    self.cards = []
    for line in lines:
      left, right = re.split(r' +\| +', re.split(': +', line)[1])
      left, right = map(int, re.split(' +', left)), map(int, re.split(' +', right))
      self.cards.append((list(left), list(right)))

  def solve_first_star(self):
    points = 0
    for winning, having in self.cards:
      matches = len(set(winning) & set(having))
      if not matches:
        continue
      points += 1 << (matches - 1)
    return points

  def solve_second_star(self):
    factors = collections.defaultdict(lambda: 1)
    count = 0
    for i, (winning, having) in enumerate(self.cards):
      count += factors[i]
      matches = len(set(winning) & set(having))
      if not matches:
        continue
      for j in range(i + 1, i + 1 + matches):
        factors[j] = factors[j] + factors[i]
    return count