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/day19.py

@@ -1,10 +1,11 @@
-import sys
-from typing import Any, Literal
+from typing import Any, Iterator, Literal
 
 import numpy as np
 import parse  # pyright: ignore[reportMissingTypeStubs]
 from numpy.typing import NDArray
 
+from ..base import BaseSolver
+
 Reagent = Literal["ore", "clay", "obsidian", "geode"]
 REAGENTS: tuple[Reagent, ...] = (
     "ore",
@@ -62,29 +63,6 @@ def dominates(lhs: State, rhs: State):
     )
 
 
-lines = sys.stdin.read().splitlines()
-
-blueprints: list[dict[Reagent, IntOfReagent]] = []
-for line in lines:
-    r: list[int] = parse.parse(  # type: ignore
-        "Blueprint {}: "
-        "Each ore robot costs {:d} ore. "
-        "Each clay robot costs {:d} ore. "
-        "Each obsidian robot costs {:d} ore and {:d} clay. "
-        "Each geode robot costs {:d} ore and {:d} obsidian.",
-        line,
-    )
-
-    blueprints.append(
-        {
-            "ore": {"ore": r[1]},
-            "clay": {"ore": r[2]},
-            "obsidian": {"ore": r[3], "clay": r[4]},
-            "geode": {"ore": r[5], "obsidian": r[6]},
-        }
-    )
-
-
 def run(blueprint: dict[Reagent, dict[Reagent, int]], max_time: int) -> int:
     # since we can only build one robot per time, we do not need more than X robots
     # of type K where X is the maximum number of K required among all robots, e.g.,
@@ -173,11 +151,31 @@ def run(blueprint: dict[Reagent, dict[Reagent, int]], max_time: int) -> int:
     return max(state.reagents["geode"] for state in state_after_t[max_time])
 
 
-answer_1 = sum(
-    (i_blueprint + 1) * run(blueprint, 24)
-    for i_blueprint, blueprint in enumerate(blueprints)
-)
-print(f"answer 1 is {answer_1}")
+class Solver(BaseSolver):
+    def solve(self, input: str) -> Iterator[Any]:
+        blueprints: list[dict[Reagent, IntOfReagent]] = []
+        for line in input.splitlines():
+            r: list[int] = parse.parse(  # type: ignore
+                "Blueprint {}: "
+                "Each ore robot costs {:d} ore. "
+                "Each clay robot costs {:d} ore. "
+                "Each obsidian robot costs {:d} ore and {:d} clay. "
+                "Each geode robot costs {:d} ore and {:d} obsidian.",
+                line,
+            )
 
-answer_2 = run(blueprints[0], 32) * run(blueprints[1], 32) * run(blueprints[2], 32)
-print(f"answer 2 is {answer_2}")
+            blueprints.append(
+                {
+                    "ore": {"ore": r[1]},
+                    "clay": {"ore": r[2]},
+                    "obsidian": {"ore": r[3], "clay": r[4]},
+                    "geode": {"ore": r[5], "obsidian": r[6]},
+                }
+            )
+
+        yield sum(
+            (i_blueprint + 1) * run(blueprint, 24)
+            for i_blueprint, blueprint in enumerate(blueprints)
+        )
+
+        yield (run(blueprints[0], 32) * run(blueprints[1], 32) * run(blueprints[2], 32))