<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>

<ul>
	<li>Every adjacent pair of words differs by a single letter.</li>
	<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
	<li><code>s<sub>k</sub> == endWord</code></li>
</ul>

<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>all the <strong>shortest transformation sequences</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]
<strong>Output:</strong> [[&quot;hit&quot;,&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;cog&quot;],[&quot;hit&quot;,&quot;hot&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]]
<strong>Explanation:</strong>&nbsp;There are 2 shortest transformation sequences:
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; &quot;cog&quot;
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;lot&quot; -&gt; &quot;log&quot; -&gt; &quot;cog&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]
<strong>Output:</strong> []
<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= beginWord.length &lt;= 5</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 500</code></li>
	<li><code>wordList[i].length == beginWord.length</code></li>
	<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
	<li><code>beginWord != endWord</code></li>
	<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
	<li>The <strong>sum</strong> of all shortest transformation sequences does not exceed <code>10<sup>5</sup></code>.</li>
</ul>

class Solution:
    def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]:
        # bidirectional BFS: construct the graph (wordList); then bfs
        wordSet = frozenset(wordList + [beginWord])
        if endWord not in wordSet:
            return []
        if beginWord == endWord:
            return [[endWord]]

        def constructGraph(wordSet: Set[str]) -> Dict[str, List[str]]:
            """
            maskedWord -> list of original words
            time: len(wordList) * len(wordList[0]) * len(wordList[0])
            space: len(wordList) * len(wordList[0]) * len(wordList[0])-letter-level
            """
            graph = collections.defaultdict(list)
            for word in wordSet:
                for i in range(len(word)):
                    key = word[:i] + '*' + word[i + 1:]  # ??? do not ignore the runtime here
                    graph[key].append(word)
            return graph

        graph = constructGraph(wordSet)
        visited = {}
        visited[beginWord] = 1
        
        def bfs(graph: Dict[str, List[str]], beginWord: str, endWord: str) -> int:
            """
            time: O(V+E) = O(len(wordList) * len(wordList[0])) * len(wordList[0])
            space: O(V) = O(len(wordList) * len(wordList[0]))
            """
            queue = collections.deque([beginWord])
            ret = 1
            while queue:
                m = len(queue)
                for _ in range(m):
                    curWord = queue.popleft()
                    if curWord == endWord:
                        return ret
                    for i in range(len(curWord)):
                        key = curWord[:i] + '*' + curWord[i + 1:]  # running time
                        for nextWord in graph[key]:
                            if nextWord not in visited:
                                visited[nextWord] = visited[curWord] + 1
                                queue.append(nextWord)
                ret += 1
            return 0

        shortDist = bfs(graph, beginWord, endWord)
        if shortDist == 0:
            return []

        ans = []
        path = []

        def backtrack(curWord: str, endWord: str) -> None:
            if curWord == endWord:
                ans.append(path + [endWord])
                return
            
            path.append(curWord)
            for i in range(len(curWord)):
                key = curWord[:i] + '*' + curWord[i + 1:]  # running time
                for nextWord in graph[key]:
                    if nextWord in visited and visited[nextWord] == visited[curWord] + 1:
                        backtrack(nextWord, endWord)
            path.pop()
        
        def backtrack2(curWord: str, beginWord: str) -> None:
            if curWord == beginWord:
                ans.append([beginWord] + list(reversed(path)))
                return
            
            path.append(curWord)
            for i in range(len(curWord)):
                key = curWord[:i] + '*' + curWord[i + 1:]  # running time
                for nextWord in graph[key]:
                    if nextWord in visited and visited[nextWord] == visited[curWord] - 1:
                        backtrack2(nextWord, beginWord)
            path.pop()

        backtrack2(endWord, beginWord)
        return ans

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