Solve 2023:02 "Cube Conundrum"

Very intermediate pythonic solution,
regex would have made the code more compact.

But since 2023:01 decreased the regex courage,
This code will do.

2023-python/output/day_02.py

@@ -0,0 +1,53 @@
+import re
+from math import prod
+from output import answer, puzzleinput
+from collections import defaultdict
+
+n = 2
+title = "Cube Conundrum"
+
+
+@answer(1, "Sum of all possible game IDs: {}")
+def part_1(data):
+    return sum(
+        [
+            id + 1
+            for id, game in enumerate(data.splitlines())
+            if not sum(
+                any(
+                    [
+                        c == "red" and int(n) > 12,
+                        c == "green" and int(n) > 13,
+                        c == "blue" and int(n) > 14,
+                    ]
+                )
+                for n, c in re.findall(r"(\d+) (\w+)", game)
+            )
+        ]
+    )
+
+
+@answer(2, "Sum of all cube set powers: {}")
+def part_2(data):
+    def power(game):
+        seen = defaultdict(int)
+        for n, c in re.findall(r"(\d+) (\w+)", game):
+            seen[c] = max([seen[c], int(n)])
+        return prod([seen["blue"], seen["red"], seen["green"]])
+
+    return sum(power(game) for game in data.splitlines())
+
+
+@puzzleinput(n)
+def parse_input(data):
+    return data
+
+
+if __name__ == "__main__":
+    parsed = parse_input()
+
+    a = part_1(parsed)
+    b = part_2(parsed)
+
+    assert a == 2439
+    assert b == 63711