Refactor code for API (#3)

Co-authored-by: Mikael CAPELLE <mikael.capelle@thalesaleniaspace.com>
Co-authored-by: Mikaël Capelle <capelle.mikael@gmail.com>
Reviewed-on: #3

src/holt59/aoc/2022/day3.py

@@ -1,23 +1,28 @@
 import string
-import sys
+from typing import Any, Iterator
 
-lines = [line.strip() for line in sys.stdin.readlines()]
+from ..base import BaseSolver
 
-# extract content of each part
-parts = [(set(line[: len(line) // 2]), set(line[len(line) // 2 :])) for line in lines]
 
-# priorities
-priorities = {c: i + 1 for i, c in enumerate(string.ascii_letters)}
+class Solver(BaseSolver):
+    def solve(self, input: str) -> Iterator[Any]:
+        lines = [line.strip() for line in input.splitlines()]
 
-# part 1
-part1 = sum(priorities[c] for p1, p2 in parts for c in p1.intersection(p2))
-print(f"answer 1 is {part1}")
+        # extract content of each part
+        parts = [
+            (set(line[: len(line) // 2]), set(line[len(line) // 2 :])) for line in lines
+        ]
 
-# part 2
-n_per_group = 3
-part2 = sum(
-    priorities[c]
-    for i in range(0, len(lines), n_per_group)
-    for c in set(lines[i]).intersection(*lines[i + 1 : i + n_per_group])
-)
-print(f"answer 2 is {part2}")
+        # priorities
+        priorities = {c: i + 1 for i, c in enumerate(string.ascii_letters)}
+
+        # part 1
+        yield sum(priorities[c] for p1, p2 in parts for c in p1.intersection(p2))
+
+        # part 2
+        n_per_group = 3
+        yield sum(
+            priorities[c]
+            for i in range(0, len(lines), n_per_group)
+            for c in set(lines[i]).intersection(*lines[i + 1 : i + n_per_group])
+        )