Merge pull request #11534 from oWnOIzRi/week307

add solution week 307 task 2 in python

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)