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Refactor code for API (#3)

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Co-authored-by: Mikael CAPELLE <mikael.capelle@thalesaleniaspace.com>
Co-authored-by: Mikaël Capelle <capelle.mikael@gmail.com>
Reviewed-on: #3
This commit is contained in:
Mikaël Capelle
2024-12-08 13:06:41 +00:00
parent ab4e3e199c
commit ce315b8778
130 changed files with 4599 additions and 3336 deletions
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34
src/holt59/aoc/2022/day2.py
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@@ -1,4 +1,6 @@
import sys
from typing import Any, Iterator
from ..base import BaseSolver
def score_1(ux: int, vx: int) -> int:
@@ -33,21 +35,23 @@ def score_2(ux: int, vx: int) -> int:
return (ux + vx - 1) % 3 + 1 + vx * 3
lines = sys.stdin.readlines()
class Solver(BaseSolver):
def solve(self, input: str) -> Iterator[Any]:
lines = input.splitlines()
# the solution relies on replacing rock / paper / scissor by values 0 / 1 / 2 and using
# modulo-3 arithmetic
#
# in modulo-3 arithmetic, the winning move is 1 + the opponent move (e.g., winning move
# if opponent plays 0 is 1, or 0 if opponent plays 2 (0 = (2 + 1 % 3)))
#
# the solution relies on replacing rock / paper / scissor by values 0 / 1 / 2 and using
# modulo-3 arithmetic
#
# in modulo-3 arithmetic, the winning move is 1 + the opponent move (e.g., winning move
# if opponent plays 0 is 1, or 0 if opponent plays 2 (0 = (2 + 1 % 3)))
#
# we read the lines in a Nx2 in array with value 0/1/2 instead of A/B/C or X/Y/Z for
# easier manipulation
values = [(ord(row[0]) - ord("A"), ord(row[2]) - ord("X")) for row in lines]
# we read the lines in a Nx2 in array with value 0/1/2 instead of A/B/C or X/Y/Z for
# easier manipulation
values = [(ord(row[0]) - ord("A"), ord(row[2]) - ord("X")) for row in lines]
# part 1 - 13526
print(f"answer 1 is {sum(score_1(*v) for v in values)}")
# part 1 - 13526
yield sum(score_1(*v) for v in values)
# part 2 - 14204
print(f"answer 2 is {sum(score_2(*v) for v in values)}")
# part 2 - 14204
yield sum(score_2(*v) for v in values)