From 6b33d543e86fc1c724d2386ec1486d4fc72326e3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?David=20Doblas=20Jim=C3=A9nez?= Date: Fri, 8 Apr 2022 16:25:18 +0200 Subject: [PATCH] Solution for problem 8 in Python --- src/Year_2018/P8.py | 55 ++++++++++++++++++++++++++++++++++++++++++++- 1 file changed, 54 insertions(+), 1 deletion(-) diff --git a/src/Year_2018/P8.py b/src/Year_2018/P8.py index da9ba78..9d09341 100644 --- a/src/Year_2018/P8.py +++ b/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()