Anders Ytterström
c0d352cdf1
I struggled a lot with the direction, but realised way too much later that the mappings are bidirectional. Part 2 was exceptionally hard, but I realised early that the order did not matter and that a simple boolean lock was all that was needed. The challenge was to know when to watch for duplicates. The seen was initially a set, but since new sets must be sent instead of references, a list is used instead to be able to use the seen + [] hack.
54 lines
1.5 KiB
Python
54 lines
1.5 KiB
Python
from solutions import BaseSolution
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from collections import defaultdict, deque, Counter
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class Solution(BaseSolution):
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input_file = "12.txt"
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def __str__(self):
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return "Day 12: Passage Pathing"
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def parse_input(self, data):
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return [l.split("-") for l in data.split()]
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def solve(self, puzzle_input):
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return self._solve(puzzle_input, lambda seen: True)
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def solve_again(self, puzzle_input):
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return self._solve(
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puzzle_input,
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lambda seen: any(
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[
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k.islower() and k != "start" and v > 1
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for k, v in Counter(seen).items()
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]
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),
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)
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def _solve(self, puzzle_input, isdup):
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ft = defaultdict(lambda: [])
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for f, t in puzzle_input:
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ft[f].append(t)
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for f, t in puzzle_input:
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ft[t].append(f)
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queue = deque([("start", list(), False)])
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s2e = 0
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while queue:
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pos, seen, dup = queue.pop()
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seen.append(pos)
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dup = isdup(seen)
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for t in ft[pos]:
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if t == "end":
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s2e += 1
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continue
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elif t == "start":
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continue
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elif dup and t.islower() and t in seen:
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continue
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queue.append((t, seen + [], dup))
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return s2e
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if __name__ == "__main__":
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solution = Solution()
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solution.show_results()
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