#!/usr/bin/python3
# -*- coding: utf-8 -*-

""" --- Day 19: Medicine for Rudolph ---

Rudolph the Red-Nosed Reindeer is sick!
His nose isn't shining very brightly, and he needs medicine.

Red-Nosed Reindeer biology isn't similar to regular reindeer biology;
Rudolph is going to need custom-made medicine.
Unfortunately, Red-Nosed Reindeer chemistry isn't similar
to regular reindeer chemistry, either.

The North Pole is equipped with a Red-Nosed Reindeer
nuclear fusion/fission plant, capable of constructing any Red-Nosed Reindeer
molecule you need. It works by starting with some input molecule
and then doing a series of replacements, one per step,
until it has the right molecule.

However, the machine has to be calibrated before it can be used.
Calibration involves determining the number of molecules that can be
generated in one step from a given starting point.

For example, imagine a simpler machine that supports
only the following replacements:

H => HO
H => OH
O => HH

Given the replacements above and starting with HOH,
the following molecules could be generated:

HOOH (via H => HO on the first H).
HOHO (via H => HO on the second H).
OHOH (via H => OH on the first H).
HOOH (via H => OH on the second H).
HHHH (via O => HH).

So, in the example above, there are 4 distinct molecules
(not five, because HOOH appears twice) after one replacement from HOH.

Santa's favorite molecule, HOHOHO, can become 7 distinct molecules
(over nine replacements: six from H, and three from O).

The machine replaces without regard for the surrounding characters.
For example, given the string H2O, the transition
H => OO would result in OO2O.

Your puzzle input describes all of the possible replacements and,
at the bottom,the medicine molecule for which you need to calibrate
the machine. How many distinct molecules can be created after all the
different ways you can do one replacement on the medicine molecule?

--- Part Two ---

Now that the machine is calibrated,
you're ready to begin molecule fabrication.

Molecule fabrication always begins with just a single electron, e,
and applying replacements one at a time, just like the ones during calibration.

For example, suppose you have the following replacements:

e => H
e => O
H => HO
H => OH
O => HH

If you'd like to make HOH, you start with e,
and then make the following replacements:

e => O to get O
O => HH to get HH
H => OH (on the second H) to get HOH

So, you could make HOH after 3 steps.
Santa's favorite molecule, HOHOHO, can be made in 6 steps.

How long will it take to make the medicine?
Given the available replacements
and the medicine molecule in your puzzle input,
what is the fewest number of steps to go from e to the medicine molecule? """


# LOGIC
# Greedy approach
# Substitute the largest substitution first,
# then continue with largest substitutions in order,
# until we reach a string

# FOR SOME REASON, THE OUTPUT FLUCTUTATES
# BETWEEN TWO VALUES THAT SEEM TO NO STABILIZE

# read input
with open('./input.txt', 'r') as f:
    text = f.readlines()


# parsed substitution as tuples of form:
# (target_string, substitution)
substitutions = {}

# parse input to get substitutions
# stop at empty line
import re
pattern = re.compile(r'(\w+) => (\w+)')
for line in text:
    if len(line.strip()) == 0:
        break
    target, sub = pattern.match(line).groups()
    substitutions[sub] = target

# add string to be changed
target_string = text[-1]
target_string = target_string.strip()

# sort the substitutions by size
# this is useful to replace the biggest strings first
substitutions_by_size = sorted(substitutions.keys(), key=len, reverse=True)


def replace(string):

    if string == 'e':
        # nothing more to replace
        return 1

    # get all applicable substitutions,
    # and then get only the first one using next
    # this is the applicable substitution because it's the largest
    try:
        substitution = next(s for s in substitutions_by_size if s in string)
    except StopIteration:
        return 1

    # replace the substitution by its molecule
    # and replace only the first occurence
    # pass this string to the replace function
    # get the number of iterations
    iterations = replace(
        string.replace(
            substitution, substitutions[substitution], 1))

    # return this + other iterations
    return 1 + iterations

print(replace(target_string))
