Solve 2023:15 "Lens Library"

WALLOFTEXT for part 2, took me 90 minutes to find
this important text:

> Each step begins with a sequence of letters that
> indicate the label of the lens on which the step
> operates. The result of running the HASH algorithm
> on the label indicates the correct box for that
> step.

It also clarifies how part 2 and part 1 relates.

2023-python/output/day_15.py

@@ -0,0 +1,63 @@
+from collections import OrderedDict, defaultdict
+
+from output import answer
+
+n = 15
+title = "Lens Library"
+
+
+@answer(1, "Sum of HASH algorithm results: {}")
+def part_1(presolved):
+    return presolved[0]
+
+
+@answer(2, "Focusing power of the resulting configuration: {}")
+def part_2(presolved):
+    return presolved[1]
+
+
+def presolve(data):
+    def h(s):
+        v = 0
+        for a in s:
+            if a == "\n":
+                continue
+            v += ord(a)
+            v *= 17
+            v = v % 256
+        return v
+
+    p1 = sum(h(c) for c in data.split(","))
+
+    b = defaultdict(OrderedDict)
+    for lr in data.split(","):
+        if "=" in lr:
+            l, r = lr.split("=")
+            if r == "":
+                continue
+            k = h(l)
+            b[k][l] = r
+        if "-" in lr:
+            l, _r = lr.split("-")
+            k = h(l)
+            if l in b[k]:
+                del b[k][l]
+    p2 = 0
+    for i, c in b.items():
+        for j, f in enumerate(b[i].values(), 1):
+            p2 += (i + 1) * j * int(f)
+
+    return p1, p2
+
+
+if __name__ == "__main__":
+    with open("./input/16.txt", "r") as f:
+        inp = f.read().strip()
+
+    inp = presolve(inp)
+
+    a = part_1(inp)
+    b = part_2(inp)
+
+    assert a == 509784
+    assert b == 230197