# 126. Word Ladder II

#### Hard

***

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]`.

&#x20;

**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.
```

&#x20;

**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**.

```python
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