This problem follows the BFS Shortest Path pattern, commonly found in the Graphs category. Recognizing this pattern is key to solving it efficiently in an interview setting.
BFS where each level changes one character. Use wildcard patterns for O(1) neighbor lookup.
BFS guarantees shortest path. Removing words from the set instead of using a visited set is cleaner and prevents re-processing.
wordSet = set(wordList)
if endWord not in wordSet: return 0
queue = deque([(beginWord, 1)])
while queue:
word, depth = queue.popleft()
for i in range(len(word)):
for c in 'abcdefghijklmnopqrstuvwxyz':
newWord = word[:i] + c + word[i+1:]
if newWord == endWord: return depth + 1
if newWord in wordSet:
wordSet.remove(newWord)
queue.append((newWord, depth + 1))
return 0Practice Word Ladder and similar Graphs problems with flashcards. Build pattern recognition through active recall.
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