#advent of code 2020
#day 18
#in my first or second aoc I coded a function to find nested brackets
#that was painful and Im not doing it again
#therefore regex

import re

def calcAdvanced(equation): #part 2
	while True:
		try:
			lb, ub = re.search(r'(\d)+\s{1}(\+){1}(\s){1}\d+', equation).span();
			subequation = equation[lb:ub].split(" ");
			equation = equation[0:lb] + str(calcOperators(subequation)) + equation[ub:];
		except:
			break;
	return int(eval(equation));
	
def calcOperators(values): #part 1
	prev = values[0];
	for v in range(2,len(values),2):
		prev = str(eval(prev+values[v-1]+values[v]));
	return prev;

def calcBrackets(eq,part):
	while True:
		try:
			lb, ub = re.search(r'\({1}([\d]|[\s]|[*]|[+])+\){1}', eq).span();
			if part == 1: 
				subeq = eq[lb+1:ub-1].split(" ");
				newval = str(calcOperators(subeq));
			else:
				subeq = eq[lb:ub];
				newval = str(calcAdvanced(subeq));
			eq = eq[0:lb] + newval + eq[ub:];
		except:
			break;
	if part == 1:
		subeq = eq.split(" ");
		return int(calcOperators(subeq));
	else: 
		return int(calcAdvanced(eq));

p1 = 0;
p2 = 0;
PuzzleInput = open("18.in","r");
for line in PuzzleInput:
	line = line[:-1];
	p1 += calcBrackets(line,1);
	p2 += calcBrackets(line,2);
PuzzleInput.close();
print("part 1 =",p1);
print("part 2 =",p2);