Solution to problem 10 part 2 in Python

src/P10.py

@@ -1,3 +1,5 @@
+from functools import lru_cache
+
 import numpy as np
 
 # --- Day 10: Adapter Array ---
@@ -120,11 +122,11 @@ import numpy as np
 with open("files/P10.txt", "r") as f:
     inputs = [int(num) for num in f.read().strip().split("\n")]
 
+# Add zero (charging outlet) and max+3 (device own built-in adapter)
+joltages = sorted(inputs + [0, max(inputs) + 3])
+
 
 def part_1() -> None:
-    # Add zero (charging outlet) and max+3 (device own built-in adapter)
-    joltages = sorted(inputs + [0, max(inputs) + 3])
-
     joltages_diffs = np.diff(joltages)
 
     print(
@@ -132,5 +134,94 @@ def part_1() -> None:
     )
 
 
+# --- Part Two ---
+
+# To completely determine whether you have enough adapters, you'll need to
+# figure out how many different ways they can be arranged. Every arrangement
+# needs to connect the charging outlet to your device. The previous rules about
+# when adapters can successfully connect still apply.
+
+# The first example above (the one that starts with 16, 10, 15) supports the
+# following arrangements:
+
+# (0), 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 6, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 5, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 6, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 6, 7, 10, 12, 15, 16, 19, (22)
+# (0), 1, 4, 7, 10, 11, 12, 15, 16, 19, (22)
+# (0), 1, 4, 7, 10, 12, 15, 16, 19, (22)
+
+# (The charging outlet and your device's built-in adapter are shown in
+# parentheses.) Given the adapters from the first example, the total number of
+# arrangements that connect the charging outlet to your device is 8.
+
+# The second example above (the one that starts with 28, 33, 18) has many
+# arrangements. Here are a few:
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 48, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 48, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 46, 49, (52)
+
+# (0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31,
+# 32, 33, 34, 35, 38, 39, 42, 45, 47, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 46, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 46, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 47, 48, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 47, 49, (52)
+
+# (0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45,
+# 48, 49, (52)
+
+# In total, this set of adapters can connect the charging outlet to your device
+# in 19208 distinct arrangements.
+
+# You glance back down at your bag and try to remember why you brought so many
+# adapters; there must be more than a trillion valid ways to arrange them!
+# Surely, there must be an efficient way to count the arrangements.
+
+# What is the total number of distinct ways you can arrange the adapters to
+# connect the charging outlet to your device?
+
+
+@lru_cache(maxsize=256)
+def recursion(node: int) -> int:
+    if node == len(joltages) - 1:
+        return 1
+    # consider last three nodes (or the end of the list, whichever is first),
+    # and for any that are valid jumps (value is no more than three more),
+    # find the number of paths from that node to the end.
+    return sum(
+        [
+            recursion(current_node)
+            for current_node in range(node + 1, min(node + 4, len(joltages)))
+            if joltages[current_node] - joltages[node] <= 3
+        ]
+    )
+
+
+def part_2() -> None:
+    total_num = recursion(0)
+    print(f"The total number of distint ways is {total_num}")
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()