Solve 2024:3 p1-2 "Mull It Over"

Felt paranoid on this one, was expecting something
much worse for pt2.

The code that solved the puzzle was _extremely_
naive:

- It asumes puzzle input only contains mul() with
1-3 chars. The versioned code have `{1,3}` as safe
guard.
- p2 asumed initial chunk does not begin with
"n't". Would have been easy to handle though.
- p2 asumed junk strings as "don", "do", "don()t"
would not exist.

In other words, I was really lucky since I did not
look that closely on puzzle input beforehand.

Might have cut 60-90 seconds further if I had just
ran pt2 immidately instead of staring dumbly on the
test data (that changed for pt2).

Also, I got reminded that \d in regular expressions
is equal to `0-9`: it does not include commas and
punctations. The code that solved the puzzle was
paranoid and instead used `0-9`.

Managed to score ~1500th somehow despite this.

2024-python/aoc.py

@@ -34,7 +34,7 @@ from collections import deque, Counter
 from heapq import heappop, heappush
 from itertools import compress, combinations, chain
 
-from output import answer  # , matrix, D, DD, ADJ, ints, mhd, mdbg, vdbg
+from output import matrix, D, DD, ADJ, ints, mhd, mdbg, vdbg
 
 
 def solve(data):

2024-python/output/__init__.py

@@ -127,6 +127,7 @@ def dijkstras(grid, start, target):
        all nodes.
     """
     import heapq
+
     target = max(grid)
     seen = set()
     queue = [(start, 0)]

2024-python/output/day_03.py

@@ -0,0 +1,26 @@
+import re
+from math import prod
+
+
+def solve(data):
+    needle = re.compile(r"mul\((\d{1,3}),(\d{1,3})\)")
+
+    p1 = sum(prod(map(int, factors)) for factors in re.findall(needle, data))
+
+    p2 = sum(
+        sum(prod(map(int, factors)) for factors in re.findall(needle, chunk))
+        for chunk in data.split("do")
+        if not chunk.startswith("n't")
+    )
+
+    return p1, p2
+
+
+if __name__ == "__main__":
+    with open("./input/03.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)