From 4a070e827b3d671be892ea9d3fd18989b28dcef8 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anders=20Engl=C3=B6f=20Ytterstr=C3=B6m?= <yttan@fastmail.se>
Date: Wed, 3 Dec 2025 07:30:04 +0100
Subject: [PATCH] Solve 2025 day 3 pt 1-2

For the first part, I used itertools.combinations
to find the highest pairs of batteries. And as
expected, that solution did not scale well for pt 2.

I figured out that reducing batteries until the top
most 12 (and 2) remained was the correct way to go.

the _maxj(line, C) function is the hive conclusion
from the solution mega thread. I really liked this
brilliant use of a while loop to exlude batteries.

 - The first char just skip the while loop. A char
   emptying the battery list also does this.
---
 2025-python/output/day_03.py | 31 +++++++++++++++++++++++++++++++
 1 file changed, 31 insertions(+)
 create mode 100644 2025-python/output/day_03.py

diff --git a/2025-python/output/day_03.py b/2025-python/output/day_03.py
new file mode 100644
index 0000000..8d993b3
--- /dev/null
+++ b/2025-python/output/day_03.py
@@ -0,0 +1,31 @@
+def solve(data):
+    p1 = 0
+    p2 = 0
+    for line in data.splitlines():
+        p1 += _maxj(line, 2)
+        p2 += _maxj(line, 12)
+    return p1, p2
+
+
+def _maxj(line, C):
+    toexcl = len(line) - C
+    batt = []
+    for c in line:
+        while toexcl and batt and batt[-1] < c:
+            toexcl -= 1
+            batt.pop()
+        batt.append(c)
+    return sum(10**x * int(y) for x, y in zip(range(C - 1, -1, -1), batt))
+
+
+if __name__ == "__main__":
+    with open("./input/03.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)
+
+    assert p1 == 17430
+    assert p2 == 171975854269367