"""
127. Word Ladder
Difficulty: Hard
https://leetcode.com/problems/word-ladder/

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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.
"""

import collections


class Solution:
    def ladderLength(self, beginWord: str, endWord: str, wordList: list[str]) -> int:
        if endWord not in wordList:
            return 0

        adj = collections.defaultdict(list)
        wordList.append(beginWord)

        for word in wordList:
            for j in range(len(word)):
                pattern = word[:j] + "*" + word[j + 1 :]
                adj[pattern].append(word)

        visit = set([beginWord])
        q = collections.deque([beginWord])
        res = 1

        while q:
            for i in range(len(q)):
                word = q.popleft()
                if word == endWord:
                    return res
                for j in range(len(word)):
                    pattern = word[:j] + "*" + word[j + 1 :]
                    for adjWord in adj[pattern]:
                        if adjWord not in visit:
                            visit.add(adjWord)
                            q.append(adjWord)
            res += 1
        return 0