13.2.py

with open('input/13.txt') as f:
    notes = [x.strip() for x in f.readlines()]
    time = int(notes[0])
    buses = [int(x.replace('x', '0')) for x in notes[1].split(',')]

def get_bus_next_journey(bus, t):
  n_trips = int(t/ bus) + 1; 
  next_trip = bus * n_trips
  return next_trip - t

def is_valid(bus, t):
  if not bus:
    return True
  return t % bus == 0

def productise(buses):
  product = 1
  for b in buses:
    if not b:
      continue
    print(b, product)
    product *= (b - buses.index(b))
  return product

print('Part Two:', productise(buses))
# Part Two: 54038417347200
# didn't work :) worth a shot as a gut instinct
# Real solution
# Part Two: 471793476184394

# start with first bus and keep track of product of buses
t = 0
p = 1

# we want the first timestamp where all buses leave in subsequent minutes
# this is equivalent to all buses leaving in the same minute if their offset
# is removed
# a bus leaves if its modulus is 0
# t % id == 0
# and a bus at index x
# t + x % id == 0
for index, bus in enumerate(buses):
  # ignore every minute buses
  if not bus:
    continue
    
  # find the next time where this bus is valid
  # step in sizes of last valid product
  # (current bus and all previous buses are valid)
  while (t + index) % bus != 0:
    t += p
  p *= bus
print('Part Two:', t)
# Part Two: 471793476184394