LeetCode - Word Ladder

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.

 

My Solution:

from collections import deque


class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        if endWord not in wordList:
            return 0
        
        def one_char_diff(word1, word2):
            for i in range(len(word1)):
                if word1[i] != word2[i]:
                    return word1[i+1:] == word2[i+1:] if i + 1 <= len(word1) - 1 else True
            
            return False

        queue = deque([(beginWord, 1)])  # (current_word, num_words)

        while queue:
            current_word, num_words = queue.popleft()
            if current_word == endWord:
                return num_words
            
            a = wordList[:]
            for word in a:
                if one_char_diff(current_word, word):
                    queue.append((word, num_words + 1))
                    wordList.remove(word)
        
        return 0