The first intuitive idea that jumps out of my mind after taking a look at LeetCode #127 is using the graph algorithm. But for building the graph first, I need to traverse the wordList by O(n2) times.

Here comes the time complexity analysis: the length of the wordList is about 5000, O(n2) means 5000*5000=25*106. For a python script in LeetCode, it will cost about 1 second for every 106 operations. Thus 25*106 will cost about 25 seconds, which is too long for a LeetCode question.

Therefore the best method to build a graph is not to traverse the wordList multiple times, but to just iterate all lower-case alphabets (be aware of the constraints beginWord, endWord, and wordList[i] consist of lowercase English letters.). By just iterating lower-case alphabets, I can reduce time to 260*5000=1.3*106 (the max length of words in wordList is 10).

The code below also uses my old trick of visited nodes.

from collections import defaultdict

class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
        words_set = set(wordList)
        conns = defaultdict(set)
        for word in wordList + [beginWord]:
            for index in range(len(word)):
                conns[word] |= {word[:index] + cand + word[index+1:] for cand in "abcdefghijklmnopqrstuvwxyz" if cand != word[index] and word[:index] + cand + word[index+1:] in words_set}
        # bfs
        queue = {beginWord}
        already = set()
        ans = 1
        while queue:
            new_queue = set()
            for node in queue:
                for _next in conns[node]:
                    if _next == endWord:
                        return ans + 1
            already |= queue
            queue = new_queue - already
            ans += 1
        return 0