126. Word Ladder II

Hard

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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 all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].

Example 1:

Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"]
Output: [["hit","hot","dot","dog","cog"],["hit","hot","lot","log","cog"]]
Explanation: There are 2 shortest transformation sequences:
"hit" -> "hot" -> "dot" -> "dog" -> "cog"
"hit" -> "hot" -> "lot" -> "log" -> "cog"

Example 2:

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

Constraints:

• 1 <= beginWord.length <= 5

• endWord.length == beginWord.length

• 1 <= wordList.length <= 1000

• wordList[i].length == beginWord.length

• beginWord, endWord, and wordList[i] consist of lowercase English letters.

• beginWord != endWord

• All the words in wordList are unique.

class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        wordList = set(wordList)
        charSet = {w for word in wordList for w in word}
        level = {}
        level[beginWord] = [[beginWord]]
        while level:
            new_level = defaultdict(list)
            for word, paths in level.items():
                if word == endWord:
                    return paths
                for index in range(len(word)):
                    for char in charSet:
                        new_word = word[:index] + char + word[index+1:]
                        if new_word in wordList:
                            for path in paths:
                                new_level[new_word].append(path + [new_word])
            wordList = wordList - set(new_level.keys())
            level = new_level
                            

Complexity : O(N*W^2)

N is number of words in wordList

W is average word length