Solution to problem 7 in Python (failed mypy :()

src/P7.py

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+from typing import Dict
+
+# --- Day 7: Handy Haversacks ---
+
+# You land at the regional airport in time for your next flight. In fact, it
+# looks like you'll even have time to grab some food: all flights are currently
+# delayed due to issues in luggage processing.
+
+# Due to recent aviation regulations, many rules (your puzzle input) are being
+# enforced about bags and their contents; bags must be color-coded and must
+# contain specific quantities of other color-coded bags. Apparently, nobody
+# responsible for these regulations considered how long they would take to
+# enforce!
+
+# For example, consider the following rules:
+
+# light red bags contain 1 bright white bag, 2 muted yellow bags.
+# dark orange bags contain 3 bright white bags, 4 muted yellow bags.
+# bright white bags contain 1 shiny gold bag.
+# muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
+# shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
+# dark olive bags contain 3 faded blue bags, 4 dotted black bags.
+# vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
+# faded blue bags contain no other bags.
+# dotted black bags contain no other bags.
+
+# These rules specify the required contents for 9 bag types. In this example,
+# every faded blue bag is empty, every vibrant plum bag contains 11 bags (5
+# faded blue and 6 dotted black), and so on.
+
+# You have a shiny gold bag. If you wanted to carry it in at least one other
+# bag, how many different bag colors would be valid for the outermost bag? (In
+# other words: how many colors can, eventually, contain at least one shiny gold
+# bag?)
+
+# In the above rules, the following options would be available to you:
+
+#     A bright white bag, which can hold your shiny gold bag directly.
+#     A muted yellow bag, which can hold your shiny gold bag directly, plus
+# some other bags.
+#     A dark orange bag, which can hold bright white and muted yellow bags,
+# either of which could then hold your shiny gold bag.
+#     A light red bag, which can hold bright white and muted yellow bags,
+# either of which could then hold your shiny gold bag.
+
+# So, in this example, the number of bag colors that can eventually contain at
+# least one shiny gold bag is 4.
+
+# How many bag colors can eventually contain at least one shiny gold bag? (The
+# list of rules is quite long; make sure you get all of it.)
+
+with open("files/P7.txt", "r") as f:
+    lines = [code for code in f.read().strip().split("\n")]
+
+# to make a dictionary of bags
+bag_types = []
+all_bags: Dict[str, str] = {}
+for line in lines:
+    main_bag = line.split(" bags contain ")[0]
+    # get a string with the content of each bag
+    contains = line.split(" bags contain ")[1:][0]
+    # house keeping
+    each_contain = contains.split(",")
+    each_contain = [cnt.lstrip() for cnt in each_contain]
+    # get a list of strings with the content of each bag
+    each_contain = [" ".join(cont.split(" ")[:-1]) for cont in each_contain]
+    # get a dictionary with bags as keys and number of them as values
+    each_contain = {
+        " ".join(cont.split(" ")[1:]): cont.split(" ")[0]
+        for cont in each_contain
+    }
+
+    if main_bag not in bag_types:
+        bag_types.append(main_bag)
+    if all_bags.get(main_bag):
+        each_contain.update(all_bags[main_bag])
+    all_bags[main_bag] = each_contain
+
+
+def check_bag(bags: Dict[str, str], my_bag: str, current_bag: str) -> int:
+    if current_bag == my_bag:
+        return 1
+    if bags.get(current_bag) is None:
+        return 0
+    else:
+        counts = []
+        for k, v in bags[current_bag].items():
+            counts.append(check_bag(bags, my_bag, k))
+        return max(counts)
+
+
+def part_1() -> None:
+    found_bags = 0
+    my_bag = "shiny gold"
+    for k, v in all_bags.items():
+        if k != my_bag:
+            found_bags += check_bag(all_bags, my_bag, k)
+    print(f"{found_bags} bags can contain a {my_bag} bag.")
+
+
+if __name__ == "__main__":
+    part_1()