From ea528b3a4f9f371f274a0c9c3012f49cb5e2bed0 Mon Sep 17 00:00:00 2001
From: Mohammad Sajid Anwar <mohammad.anwar@yahoo.com>
Date: Wed, 5 Feb 2025 11:08:16 +0000
Subject: - Added solutions by Steven Wilson.

---
 challenge-307/steven-wilson/python/ch-02.py | 39 -----------------------------
 challenge-307/steven-wilson/python/ch-2.py  | 39 +++++++++++++++++++++++++++++
 2 files changed, 39 insertions(+), 39 deletions(-)
 delete mode 100644 challenge-307/steven-wilson/python/ch-02.py
 create mode 100644 challenge-307/steven-wilson/python/ch-2.py

diff --git a/challenge-307/steven-wilson/python/ch-02.py b/challenge-307/steven-wilson/python/ch-02.py
deleted file mode 100644
index f910e3bf29..0000000000
--- a/challenge-307/steven-wilson/python/ch-02.py
+++ /dev/null
@@ -1,39 +0,0 @@
-#!/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)
diff --git a/challenge-307/steven-wilson/python/ch-2.py b/challenge-307/steven-wilson/python/ch-2.py
new file mode 100644
index 0000000000..f910e3bf29
--- /dev/null
+++ b/challenge-307/steven-wilson/python/ch-2.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)
-- 
cgit