From bf4373ced86010452a5c87f2aea9ec7d1ccd53cf Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Tue, 1 Mar 2022 10:46:32 +0100
Subject: [PATCH] Solution to problem 5 in Python

---
 src/Year_2016/P5.py | 58 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 58 insertions(+)

diff --git a/src/Year_2016/P5.py b/src/Year_2016/P5.py
index ecaf05b..46f7417 100644
--- a/src/Year_2016/P5.py
+++ b/src/Year_2016/P5.py
@@ -30,7 +30,9 @@
 # Your puzzle input is abbhdwsy.
 
 
+from collections import defaultdict
 from hashlib import md5
+from typing import DefaultDict
 
 _input = "abbhdwsy"
 
@@ -48,5 +50,61 @@ def part_1() -> None:
             break
 
 
+# --- Part Two ---
+
+# As the door slides open, you are presented with a second door that uses a
+# slightly more inspired security mechanism. Clearly unimpressed by the last
+# version (in what movie is the password decrypted in order?!), the Easter
+# Bunny engineers have worked out a better solution.
+
+# Instead of simply filling in the password from left to right, the hash now
+# also indicates the position within the password to fill. You still look for
+# hashes that begin with five zeroes; however, now, the sixth character
+# represents the position (0-7), and the seventh character is the character to
+# put in that position.
+
+# A hash result of 000001f means that f is the second character in the
+# password. Use only the first result for each position, and ignore invalid
+# positions.
+
+# For example, if the Door ID is abc:
+
+#     The first interesting hash is from abc3231929, which produces 0000015...;
+# so, 5 goes in position 1: _5______.
+#     In the previous method, 5017308 produced an interesting hash; however, it
+# is ignored, because it specifies an invalid position (8).
+#     The second interesting hash is at index 5357525, which produces
+# 000004e...; so, e goes in position 4: _5__e___.
+
+# You almost choke on your popcorn as the final character falls into place,
+# producing the password 05ace8e3.
+
+# Given the actual Door ID and this new method, what is the password? Be extra
+# proud of your solution if it uses a cinematic "decrypting" animation.
+
+
+def part_2() -> None:
+    passwd = defaultdict(list)
+    end = "9" * len(_input)
+    for num in range(int(end)):
+        testing = _input + str(num)
+        result = md5(testing.encode()).hexdigest()
+        if (
+            result.startswith("0" * 5)
+            # only numbers are allow
+            and str(result[5]).isdecimal()
+            # the passwd is 8 letters long
+            and int(result[5]) < 8
+        ):
+            passwd[str(result[5])].append(result[6])
+        if len(passwd.keys()) == 8:
+            ordered_passwd = dict(sorted(passwd.items()))
+            # take the first result for each key
+            my_lst = [elem[0] for elem in ordered_passwd.values()]
+            print(f"The password is {''.join(my_lst)}")
+            break
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()