0127.Word Ladder

Link: 127. Word Ladder - Hard
Track: Amazon Tag

Question

Restate

Edge Case

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Method 1
Method 2

Method

Approach

Complexity

• Time Complexity: O($M\ast N\ast 26\ast L$)
• Space Complexity: O(N)

Code

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        wordList_set = set(wordList)
        if endWord not in wordList_set:
            return 0
        if beginWord == endWord:
            return 1
        
        queue = deque([beginWord])
        visited = set([beginWord])
        count = 1
 
        while queue:
            for _ in range(len(queue)):
                word = queue.popleft()
                
                for i in range(len(word)): # O(L)
                    for c in 'abcdefghijklmnopqrstuvwxyz':
                        new_word = word[:i] + c + word[i + 1:] # O(L)
                        if new_word == endWord:
                            return count + 1
            
                        if (new_word in wordList_set 
                            and new_word not in visited):
                            queue.append(new_word)
                            visited.add(new_word)
            count += 1
        return 0

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History

• Jan-xx-2026 Peeked