[py] Port 2015_05

py/aoc/__init__.py

@@ -2,7 +2,7 @@ import sys
 import argparse
 from pathlib import Path
 
-from .y2015 import d01, d02, d03, d04
+from .y2015 import d01, d02, d03, d04, d05
 from .y2016 import d01, d02, d03, d04, d05
 from .y2017 import d01, d02, d03, d04, d05, d06, d07, d08, d09
 from .y2018 import d01, d02, d03, d04, d05, d06, d07, d08, d09, d10, d11
@@ -15,6 +15,7 @@ DAYS = {
     "2015_02": y2015.d02.solve,
     "2015_03": y2015.d03.solve,
     "2015_04": y2015.d04.solve,
+    "2015_05": y2015.d05.solve,
     "2016_01": y2016.d01.solve,
     "2016_02": y2016.d02.solve,
     "2016_03": y2016.d03.solve,

py/aoc/y2015/d05.py

@@ -0,0 +1,56 @@
+def load_words(inputstr):
+    words = []
+    for line in inputstr.splitlines():
+        words.append(line)
+    return words
+
+
+# PART 1
+
+
+def is_vowel(char):
+    return char in "aeiou"
+
+
+def has_double_letter(word):
+    # This would look nicer im haskell:
+    return count(zip(word, word[1:]), lambda x: x[0] == x[1]) > 0
+
+
+def is_nice(word):
+    if "ab" in word or "cd" in word or "pq" in word or "xy" in word:
+        return False
+    if len(list(filter(is_vowel, word))) < 3:
+        return False
+    return has_double_letter(word)
+
+
+def count(what, function):
+    # This would also look nicer in haskell
+    return len(list(filter(function, what)))
+
+
+# PART 2
+
+
+def has_pair(word):
+    for i in range(len(word)):
+        if word[i : i + 2] in word[i + 2 :]:
+            return True
+    return False
+
+
+def has_repeat_letter(word):
+    return count(zip(word, word[2:]), lambda x: x[0] == x[1]) > 0
+
+
+def is_nice_2(word):
+    return has_pair(word) and has_repeat_letter(word)
+
+
+def solve(inputstr):
+    words = load_words(inputstr)
+    amount = count(words, is_nice)
+    print(f"Part 1: {amount}")
+    amount_2 = count(words, is_nice_2)
+    print(f"Part 2: {amount_2}")