Solution for problem 8 in Python

src/Year_2018/P8.py

@@ -47,6 +47,7 @@
 
 # What is the sum of all metadata entries?
 
+from copy import copy
 from typing import List
 
 with open("files/P8.txt") as f:
@@ -65,8 +66,60 @@ def sum_metadata(lst: List[int]) -> int:
 
 
 def part_1() -> None:
-    print(f"The sum of all metadata entries is {sum_metadata(tree)}")
+    tree_p1 = copy(tree)
+    print(f"The sum of all metadata entries is {sum_metadata(tree_p1)}")
+
+
+# --- Part Two ---
+
+# The second check is slightly more complicated: you need to find the value of
+# the root node (A in the example above).
+
+# The value of a node depends on whether it has child nodes.
+
+# If a node has no child nodes, its value is the sum of its metadata entries.
+# So, the value of node B is 10+11+12=33, and the value of node D is 99.
+
+# However, if a node does have child nodes, the metadata entries become indexes
+# which refer to those child nodes. A metadata entry of 1 refers to the first
+# child node, 2 to the second, 3 to the third, and so on. The value of this
+# node is the sum of the values of the child nodes referenced by the metadata
+# entries. If a referenced child node does not exist, that reference is
+# skipped. A child node can be referenced multiple time and counts each time it
+# is referenced. A metadata entry of 0 does not refer to any child node.
+
+# For example, again using the above nodes:
+
+#     Node C has one metadata entry, 2. Because node C has only one child node,
+# 2 references a child node which does not exist, and so the value of node C is
+# 0.
+#     Node A has three metadata entries: 1, 1, and 2. The 1 references node A's
+# first child node, B, and the 2 references node A's second child node, C.
+# Because node B has a value of 33 and node C has a value of 0, the value of
+# node A is 33+33+0=66.
+
+# So, in this example, the value of the root node is 66.
+
+# What is the value of the root node?
+
+
+def find_root_value(lst: List[int]) -> int:
+    childs = lst.pop(0)
+    metadata = lst.pop(0)
+    values = [find_root_value(lst) for _ in range(childs)]
+    metadatas = [lst.pop(0) for _ in range(metadata)]
+    if childs == 0:
+        return sum(metadatas)
+    return sum(
+        values[child - 1] for child in metadatas if child - 1 in range(childs)
+    )
+
+
+def part_2() -> None:
+    tree_p2 = copy(tree)
+    print(f"The value of the root node is {find_root_value(tree_p2)}")
 
 
 if __name__ == "__main__":
     part_1()
+    part_2()