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submitted 2 weeks ago* (last edited 2 weeks ago) by CameronDev@programming.dev to c/advent_of_code@programming.dev

Day 3: Mull It Over

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[-] zarlin@lemmy.world 2 points 2 weeks ago

Nim

import ../aoc, re, sequtils, strutils, math

proc mulsum*(line:string):int=
  let matches = line.findAll(re"mul\([0-9]{1,3},[0-9]{1,3}\)")
  let pairs = matches.mapIt(it[4..^2].split(',').map(parseInt))
  pairs.mapIt(it[0]*it[1]).sum

proc filter*(line:string):int=
  var state = true;
  var i=0
  while i < line.len:
    if state:
      let off = line.find("don't()", i)
      if off == -1:
        break
      result += line[i..<off].mulsum
      i = off+6
      state = false
    else:
      let on = line.find("do()", i)
      if on == -1:
        break
      i = on+4
      state = true
      
  if state:
    result += line[i..^1].mulsum

proc solve*(input:string): array[2,int] =
  #part 1&2
  result = [input.mulsum, input.filter]

I had a nicer solution in mind for part 2, but for some reason nre didn't want to work for me, and re couldn't give me the start/end or all results, so I ended up doing this skip/toggle approach.

Also initially I was doing it line by line out of habit from other puzzles, but then ofc the don't()s didn't propagate to the next line.

this post was submitted on 03 Dec 2024
20 points (95.5% liked)

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