From 4c3960b56e872c46346d933a850bca2ebac98854 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Mon, 13 Jun 2022 21:04:02 +0200
Subject: [PATCH] Solution to problem 11 in Python

---
 src/Year_2015/P11.py | 105 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 105 insertions(+)
 create mode 100644 src/Year_2015/P11.py

diff --git a/src/Year_2015/P11.py b/src/Year_2015/P11.py
new file mode 100644
index 0000000..4d93578
--- /dev/null
+++ b/src/Year_2015/P11.py
@@ -0,0 +1,105 @@
+# --- Day 11: Corporate Policy ---
+
+# Santa's previous password expired, and he needs help choosing a new one.
+
+# To help him remember his new password after the old one expires, Santa has
+# devised a method of coming up with a password based on the previous one.
+# Corporate policy dictates that passwords must be exactly eight lowercase
+# letters (for security reasons), so he finds his new password by incrementing
+# his old password string repeatedly until it is valid.
+
+# Incrementing is just like counting with numbers: xx, xy, xz, ya, yb, and so
+# on. Increase the rightmost letter one step; if it was z, it wraps around to
+# a, and repeat with the next letter to the left until one doesn't wrap around.
+
+# Unfortunately for Santa, a new Security-Elf recently started, and he has
+# imposed some additional password requirements:
+
+#     Passwords must include one increasing straight of at least three letters,
+# like abc, bcd, cde, and so on, up to xyz. They cannot skip letters; abd
+# doesn't count.
+#     Passwords may not contain the letters i, o, or l, as these letters can be
+# mistaken for other characters and are therefore confusing.
+#     Passwords must contain at least two different, non-overlapping pairs of
+# letters, like aa, bb, or zz.
+
+# For example:
+
+#     hijklmmn meets the first requirement (because it contains the straight
+# hij) but fails the second requirement requirement (because it contains i and
+# l).
+#     abbceffg meets the third requirement (because it repeats bb and ff) but
+# fails the first requirement.
+#     abbcegjk fails the third requirement, because it only has one double
+# letter (bb).
+#     The next password after abcdefgh is abcdffaa.
+#     The next password after ghijklmn is ghjaabcc, because you eventually skip
+# all the passwords that start with ghi..., since i is not allowed.
+
+# Given Santa's current password (your puzzle input), what should his next
+# password be?
+
+import re
+from string import ascii_lowercase
+
+with open("files/P11.txt", "r") as f:
+    password = [c for c in f.read().strip()]
+
+forbidden_chars = {105, 108, 111}
+
+
+def rule_1(chars: list[str]) -> bool:
+    # increasing straight of at least three letters
+    return any(
+        "".join(chars[i : i + 3]) in ascii_lowercase
+        for i in range(len(chars) - 2)
+    )
+
+
+def rule_2(chars: list[str], pos: int) -> list[str]:
+    # not contain the letters i, o, or l
+    a = ord(chars[pos]) + 1
+    # rule # 2
+    if a in forbidden_chars:
+        a += 1
+    # wraps around to a
+    if a > 122:
+        a = 97
+        chars[pos] = chr(a)
+        rule_2(chars, pos - 1)
+    else:
+        chars[pos] = chr(a)
+    return chars
+
+
+def rule_3(chars: list[str]) -> bool:
+    # at least two different, non-overlapping pairs of letters
+    return bool(re.search(r"(\w)\1.*(\w)\2", "".join(chars)))
+
+
+def part_1() -> None:
+    new_password = password
+    found = False
+    while not found:
+        new_password = rule_2(new_password, -1)
+        # rule 1
+        if not rule_1(new_password):
+            continue
+        # rule # 3
+        if not rule_3(new_password):
+            continue
+
+        res = "".join(new_password)
+        print(f"Santa's next password is {res}")
+        found = True
+
+
+# --- Part Two ---
+
+# Santa's password expired again. What's the next one?
+
+
+if __name__ == "__main__":
+    part_1()
+    # Same code again
+    part_1()