Advent of code 2020 day 10

def product_of_number_of_differences(s):
  l = sorted(map(int, s.split()))
  counter = collections.Counter(np.diff([0] + l + [l[-1] + 3]))
  return counter[1] * counter[3]


# Version 1; exotic.
def count_combinations(s):
  diff = np.diff([0] + sorted(map(int, s.split())))
  lengths_of_ones = map(len, re.findall('1+', ''.join(map(str, diff))))

  def f(n):
    """Number of combinations from a sequence of n consecutive one-diffs."""
    return 0 if n < 0 else 1 if n < 2 else f(n - 1) + f(n - 2) + f(n - 3)

  return np.prod(list(map(f, lengths_of_ones)))


# Version 2; inspired by Ray's solution.
def count_combinations(s):
  l = [0] + sorted(map(int, s.split()))
  npaths = [1] + [0] * (len(l) - 1)
  for i in range(len(l)):
    for j in range(i + 1, min(i + 4, len(l))):
      if l[j] - l[i] <= 3:
        npaths[j] += npaths[i]
  return npaths[-1]


# Version 3; "gather/pull".
def count_combinations(s):
  l = [0] + sorted(map(int, s.split()))
  npaths = [1]
  for i in range(1, len(l)):
    npaths.append(
        sum(npaths[j]
            for j in range(-4, 0)
            if i + j >= 0 and l[i] - l[i + j] < 4))
  return npaths[-1]