Solution to problem 4 in Python

src/Year_2018/P4.py

@@ -91,7 +91,7 @@ with open("files/P4.txt") as f:
 sorted_schedule = sorted(schedule)
 
 
-def part_1() -> None:
+def part_1():
     schedule_dict = defaultdict(list)
     for line in sorted_schedule:
         split_str = re.split(r"] | #| |\[", line)
@@ -124,10 +124,36 @@ def part_1() -> None:
     # get the most common minute for the guard who sleep the most
     most_common_min = most_common_dict[max_sleep_length][0][0]
 
-    print(
+    print(f"Result (part 1): {int(max_sleep_length) * int(most_common_min)}")
-        f"The result we are looking for is {int(max_sleep_length) * int(most_common_min)}"
+
-    )
+    return most_common_dict
+
+
+# --- Part Two ---
+
+# Strategy 2: Of all guards, which guard is most frequently asleep on the same
+# minute?
+
+# In the example above, Guard #99 spent minute 45 asleep more than any other
+# guard or minute - three times in total. (In all other cases, any guard spent
+# any minute asleep at most twice.)
+
+# What is the ID of the guard you chose multiplied by the minute you chose?
+# (In the above example, the answer would be 99 * 45 = 4455.)
+
+
+def part_2():
+    most_common_dict = part_1()
+
+    # get most frequently asleep minute for any guard
+    most_frequent_min = max(v[0][1] for v in most_common_dict.values())
+
+    for k, v in most_common_dict.items():
+        # get the key for that particular value
+        if most_frequent_min in v[0]:
+            print(f"Result (part 2): {int(k) * int(v[0][0])}")
 
 
 if __name__ == "__main__":
-    part_1()
+    # part_1()
+    part_2()