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| -rw-r--r-- | challenge-307/steven-wilson/python/ch-02.py | 39 |
1 files changed, 39 insertions, 0 deletions
diff --git a/challenge-307/steven-wilson/python/ch-02.py b/challenge-307/steven-wilson/python/ch-02.py new file mode 100644 index 0000000000..f910e3bf29 --- /dev/null +++ b/challenge-307/steven-wilson/python/ch-02.py @@ -0,0 +1,39 @@ +#!/usr/bin/env python3 + + +def find_anagrams(*words): + """ Given a list of words, find any two consecutive words and if they are + anagrams, drop the first word and keep the second. You continue this until + there is no more anagrams in the given list and return the count of final + list. + >>> find_anagrams("acca", "dog", "god", "perl", "repl") + 3 + >>> find_anagrams("abba", "baba", "aabb", "ab", "ab") + 2 + """ + length_words = len(words) + drop = 0 + + if length_words < 2: + raise ValueError("At least 2 arguments are required") + + for n, _ in enumerate(words): + if n < length_words - 1 and are_anagrams(words[n], words[n+1]): + drop += 1 + return length_words - drop + + +def are_anagrams(first, second): + """ Are two words an anagram of each other + >>> are_anagrams("dog", "god") + True + >>> are_anagrams("aabb", "ab") + False + """ + return sorted(first) == sorted(second) + + +if __name__ == "__main__": + import doctest + + doctest.testmod(verbose=True) |