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authorSteven <steven1170@zoho.eu>2025-02-04 20:55:43 +0000
committerSteven <steven1170@zoho.eu>2025-02-04 20:55:43 +0000
commit735978e34a40c558e7e3996d7ac575eed6ea4738 (patch)
treef8131858102a0ba784816649c7a3a156606c0e13
parent9e5deef559ffa5bb3cab2a89475de0c0a370c59b (diff)
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add solution week 307 task 2 in python
-rw-r--r--challenge-307/steven-wilson/python/ch-02.py39
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
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+++ b/challenge-307/steven-wilson/python/ch-02.py
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+#!/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)