Add Python solution for Word Ladder (practice256)

solutions/python/practice256.py

@@ -1,11 +1,32 @@
 # Problem: Word Ladder
 # Source: https://leetcode.com/problems/word-ladder/
-# Note: Finds the length of the shortest transformation sequence between two words, changing one letter at a time using dictionary words.
-# Status: 🔄 IN PROGRESS - Implementation needed
-# TODO: Implement BFS for shortest transformation sequence
+# Note: Implement BFS for shortest transformation sequence using deque
+# Status: ✅ Done
 
-def word_ladder(word):
-    pass
+from collections import deque
 
-dictionary = {"hot", "dot", "dog", "lot", "log", "cog"}
+def word_ladder(beginWord, endWord, wordList):
+    wordSet = set(wordList)
+    if endWord not in wordSet: 
+        return 0
+    queue = deque([(beginWord, 1)])
+    wordSet.discard(beginWord)
+
+    # BFS
+    while queue: 
+        word, steps = queue.popleft()
+        if word == endWord: 
+            return steps
+        for i in range(len(word)): 
+            for character in "abcdefghijklmnopqrstuvwxyz": 
+                if character == word[i]: 
+                    continue
+                newWord = word[:i] + character + word[i+1:]
+                if newWord in wordSet: 
+                    wordSet.remove(newWord)
+                    queue.append((newWord, steps + 1))
+    return 0
+
+# Example
+dictionary = ["hot", "dot", "dog", "lot", "log", "cog"]
 print(word_ladder("hit", "cog", dictionary))  # 5