[py] Simplify 2015_05

py/aoc/y2015/d05.py

@@ -1,56 +1,20 @@
-def load_words(inputstr):
+import re
-    words = []
-    for line in inputstr.splitlines():
-        words.append(line)
-    return words
 
 
-# PART 1
+def part1_nice(word):
+    vowels = re.search(r"([aeiou].*){3}", word)
+    double = re.search(r"(.)\1", word)
+    forbidden = re.search(r"ab|cd|pq|xy", word)
+    return vowels and double and not forbidden
 
 
-def is_vowel(char):
+def part2_nice(word):
-    return char in "aeiou"
+    pair = re.search(r"(..).*\1", word)
-
+    repeats = re.search(r"(.).\1", word)
-
+    return pair and repeats
-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)
+    words = inputstr.splitlines()
-    amount = count(words, is_nice)
+    print(f"Part 1: {len(list(filter(part1_nice, words)))}")
-    print(f"Part 1: {amount}")
+    print(f"Part 2: {len(list(filter(part2_nice, words)))}")
-    amount_2 = count(words, is_nice_2)
-    print(f"Part 2: {amount_2}")