Nim

Easy one today. Part 2 is pretty forgiving on performance, so regex bruteforce was only a couple seconds . But eventually I've cleaned it up and did a solution that runs in ~340 ms.

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

proc isRepeating(str:string, sectorLength=1): bool =
  if str.len mod sectorLength != 0: return false
  for i in countUp(0, str.len - sectorLength, sectorLength):
    if str.toOpenArray(i, i+sectorLength-1) != str.toOpenArray(0, sectorLength-1):
      return false
  true

proc solve(input: string): AOCSolution[int, int] =
  let ranges = input.split(',').mapIt:
    let parts = it.split('-')
    (parseInt parts[0], parseInt parts[1])

  for (a, b) in ranges:
    for num in a .. b:
      if num < 10: continue
      let strnum = $num
      let half = strnum.len div 2

      for i in countDown(half, 1):
        if strnum.isRepeating(i):
          if i == half and strnum.len mod 2 == 0:
            result.part1 += num
          result.part2 += num
          break

Full solution at Codeberg: solution.nim