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127. Word Ladder

Difficulty Topics

Description

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

 

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long.

Example 2:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore there is no valid transformation sequence.

 

Constraints:

  • 1 <= beginWord.length <= 10
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 5000
  • wordList[i].length == beginWord.length
  • beginWord, endWord, and wordList[i] consist of lowercase English letters.
  • beginWord != endWord
  • All the words in wordList are unique.

Solution

word-ladder.py
class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        n = len(beginWord)
        wordList = set(wordList)
        visited = set([beginWord])
        if endWord not in wordList: return 0

        queue = deque([(beginWord, 1)])

        while queue:
            word, steps = queue.popleft()

            if word == endWord: return steps

            for i in range(n):
                for c in string.ascii_lowercase:
                    new = word[:i] + c + word[i + 1:]
                    if new in wordList and new not in visited:
                        visited.add(new)
                        queue.append((new, steps + 1))

        return 0