Solve 2024:19 p1-2 "Linen Layout"

Initial tries to use a while loop instead of
recursion gave a classic AoC situation: test cases
worked, but not actual AoC puzzle input.

Turns out I only considererd removing the longest
pattern form the design, rather than consider all
possible removals.

Some Python goodies in here:

- `"abcdefgh".startswith("abc")` instead of regexp.
- removeprefix() is nice, and in some cases more
readable. `"abcdefgh".removeprefix("abc")` vs
`"abcdefgh[3:]"
- To speed things up for pt 2, functools is used
which requires a dict to be hashed as a tuple.

If one wish to solve this without recursion, a
BFS solution is most likely the way to go.

def ispossible(design, patterns):
    Q = [design]
    possible = 0
    while Q:
        remaining = Q.pop(0)
        if not remaining:
            possible += 1
            continue
        for pattern in patterns[remaining[0]]:
            if remaining.startswith(pattern):
                Q.append(remaining.removeprefix(pattern))
    return possible

2024-python/output/day_19.py

@@ -0,0 +1,39 @@
+from functools import cache
+from collections import defaultdict
+
+
+def solve(data):
+    patterns, designs = data.split("\n\n")
+    patterns = sorted(patterns.split(", "), key=lambda s: -len(s))
+    designs = designs.split()
+
+    hashed = defaultdict(list)
+    for p in patterns:
+        hashed[p[0]].append(p)
+    hashed = tuple([(k, tuple(v)) for k, v in hashed.items()])
+
+    possible = sum(ispossible(d, hashed) > 0 for d in designs)
+    every_possibility = sum(ispossible(d, hashed) for d in designs)
+    return possible, every_possibility
+
+
+@cache
+def ispossible(remaining, patterns):
+    if not remaining:
+        return 1
+    selected = dict(patterns).get(remaining[0], [])
+    return sum(
+        ispossible(remaining.removeprefix(pattern), patterns)
+        for pattern in selected
+        if remaining.startswith(pattern)
+    )
+
+
+if __name__ == "__main__":
+    with open("./input/19.txt", "r") as f:
+        inp = f.read().strip()
+
+    p1, p2 = solve(inp)
+
+    print(p1)
+    print(p2)