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#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);
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