Solution to problem 7 part 2 in Python (failed mypy:()

src/P7.py

@@ -98,5 +98,71 @@ def part_1() -> None:
     print(f"{found_bags} bags can contain a {my_bag} bag.")
 
 
+# --- Part Two ---
+
+# It's getting pretty expensive to fly these days - not because of ticket
+# prices, but because of the ridiculous number of bags you need to buy!
+
+# Consider again your shiny gold bag and the rules from the above example:
+
+#     faded blue bags contain 0 other bags.
+#     dotted black bags contain 0 other bags.
+#     vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted
+# black bags.
+#     dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted
+# black bags.
+
+# So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags
+# within it) plus 2 vibrant plum bags (and the 11 bags within each of those):
+# 1 + 1*7 + 2 + 2*11 = 32 bags!
+
+# Of course, the actual rules have a small chance of going several levels
+# deeper than this example; be sure to count all of the bags, even if the
+# nesting becomes topologically impractical!
+
+# Here's another example:
+
+# shiny gold bags contain 2 dark red bags.
+# dark red bags contain 2 dark orange bags.
+# dark orange bags contain 2 dark yellow bags.
+# dark yellow bags contain 2 dark green bags.
+# dark green bags contain 2 dark blue bags.
+# dark blue bags contain 2 dark violet bags.
+# dark violet bags contain no other bags.
+
+# In this example, a single shiny gold bag must contain 126 other bags.
+
+# How many individual bags are required inside your single shiny gold bag?
+
+
+bags_contains: Dict[str, str] = {}
+for k, v in all_bags.items():
+    bags_contains[k] = []
+    try:
+        for kk, vv in v.items():
+            bags_contains[k] += [kk] * int(vv)
+    except ValueError:
+        pass
+
+
+def count_bags(current_bag: str) -> int:
+    if current_bag == " " or bags_contains.get(current_bag) is None:
+        return 0
+
+    cnt = len(bags_contains[current_bag])
+    nbags = []
+    for k in bags_contains[current_bag]:
+        nbags.append(count_bags(k))
+    return sum(nbags) + cnt
+
+
+def part_2() -> None:
+    my_bag = "shiny gold"
+    nbags = count_bags(my_bag)
+
+    print(f"My shiny gold bag can contain {nbags} bags")
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()