Anders Ytterström
99f5385f25
Yes, I do "".join(). Python is not always beautiful. I began reusing part 1 for part 2, until I realised you cannot reuse the Counter.most_common, and you need the actual values to be able to se equal occourences. I probably lost 5-15 minutes just to dribble with 3 levels of nested objects. In GMT+1 before coffee, that cost me. Part 2 was way uglier before some well motivated refactoring. Since all tests and expected output were in place, refactoring was easy.
43 lines
1.2 KiB
Python
43 lines
1.2 KiB
Python
from collections import Counter
|
|
from solutions import BaseSolution
|
|
|
|
|
|
class Solution(BaseSolution):
|
|
input_file = "03.txt"
|
|
|
|
def __str__(self):
|
|
return "Day 3: Binary Diagnostic"
|
|
|
|
def parse_input(self, data):
|
|
return data.split()
|
|
|
|
def solve(self, puzzle_input):
|
|
e, g = self._eg(puzzle_input)
|
|
return int("".join(g), 2) * int("".join(e), 2)
|
|
|
|
def solve_again(self, puzzle_input):
|
|
og = self._ogco2(puzzle_input, "1")
|
|
co2 = self._ogco2(puzzle_input, "0")
|
|
return int("".join(og), 2) * int("".join(co2), 2)
|
|
|
|
def _eg(self, puzzle_input):
|
|
e = []
|
|
g = []
|
|
for r in zip(*puzzle_input):
|
|
x, y = Counter(r).most_common()
|
|
e.append(x[0])
|
|
g.append(y[0])
|
|
return e, g
|
|
|
|
def _ogco2(self, values, default, i=0):
|
|
eg = [Counter(r).most_common() for r in zip(*values)]
|
|
k = default if eg[i][0][1] == eg[i][1][1] else eg[i][abs(int(default) - 1)][0]
|
|
rem = [v for v in values if v[i] == k]
|
|
if len(rem) == 1:
|
|
return rem
|
|
return self._ogco2(rem, default, i + 1)
|
|
|
|
|
|
if __name__ == "__main__":
|
|
solution = Solution()
|
|
solution.show_results()
|