Advent of code 2025 day 2 super-fast

# Arithmetic solution.  E.g., a 6-digit value divisible by 111111, 10101, or 1001 consists of
# a repeated digit sequence (of length 1, 2, or 3).  We iterate over the in-range multiples.

def day2(s, *, part2=False):

  def split_by_num_digits(start, stop):
    """Partition an integer range into subranges with the same number of digits."""
    while start < stop:
      end = min(10 ** (math.floor(math.log10(start)) + 1), stop)
      yield start, end
      start = end

  total = 0
  for lo, hi in re.findall(r'(\d+)-(\d+)', s):
    for start, stop in split_by_num_digits(int(lo), int(hi) + 1):
      num_digits = math.floor(math.log10(start)) + 1
      divisors = [
          (10**num_digits - 1) // (10**p - 1)
          for p in range(1, num_digits)
          if num_digits % p == 0 and (p * 2 == num_digits or part2)
      ]
      values = set()
      for d in divisors:
        values.update(range(-(-start // d) * d, (stop - 1) // d * d + 1, d))
      total += sum(values)

  return total