Day 7: Laboratories

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[-] janAkali@lemmy.sdf.org 3 points 2 months ago* (last edited 2 months ago)

Nim

Another simple one.

Part 1: count each time a beam crosses a splitter.
Part 2: keep count of how many particles are in each column in all universes
(e.g. with a simple 1d array), then sum.

Runtime: ~~116 μs~~ ~~95 µs~~ 86 µs

old version

type
  AOCSolution[T,U] = tuple[part1: T, part2: U]

proc solve(input: string): AOCSolution[int, int] =
  var beams = newSeq[int](input.find '\n')
  beams[input.find 'S'] = 1

  for line in input.splitLines():
    var newBeams = newSeq[int](beams.len)
    for pos, cnt in beams:
      if cnt == 0: continue
      if line[pos] == '^':
        newBeams[pos-1] += cnt
        newBeams[pos+1] += cnt
        inc result.part1
      else:
        newbeams[pos] += cnt
    beams = newBeams
  result.part2 = beams.sum()

Update: found even smaller and faster version that only needs a single array.
Update #2: small optimization

type
  AOCSolution[T,U] = tuple[part1: T, part2: U]

proc solve(input: string): AOCSolution[int, int] =
  var beams = newSeq[int](input.find '\n')
  beams[input.find 'S'] = 1

  for line in input.splitLines():
    for pos, c in line:
      if c == '^' and beams[pos] > 0:
        inc result.part1
        beams[pos-1] += beams[pos]
        beams[pos+1] += beams[pos]
        beams[pos] = 0
  result.part2 = beams.sum()

Full solution at Codeberg: solution.nim

[-] Deebster@programming.dev 2 points 2 months ago

Ah, it took me looking at your updated Codeberg version to understand this - you looked at part two in the opposite way than I did but it comes out the same in the end (and yours is much more efficient).

this post was submitted on 07 Dec 2025

32 points (97.1% liked)

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