<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 {
public:
    unordered_map<string, vector<string>> adjList;
    vector<string> currPath;
    vector<vector<string>> shortestPaths;

    vector<string> findNeighbors(string &word,
                                 unordered_set<string> &wordList) {
        vector<string> neighbors;

        for (int i = 0; i < word.size(); i++) {
            char oldChar = word[i];

            // replace the i-th character with all letters from a to z except
            // the original character
            for (char c = 'a'; c <= 'z'; c++) {
                word[i] = c;

                // skip if the character is same as original or if the word is
                // not present in the wordList
                if (c == oldChar || !wordList.count(word)) {
                    continue;
                }
                neighbors.push_back(word);
            }
            word[i] = oldChar;
        }
        return std::move(neighbors);
    }

    void backtrack(string &source, string &destination) {
        // store the path if we reached the endWord
        if (source == destination) {
            shortestPaths.push_back(
                vector<string>(currPath.rbegin(), currPath.rend()));
        }
        for (int i = 0; i < adjList[source].size(); i++) {
            currPath.push_back(adjList[source][i]);
            backtrack(adjList[source][i], destination);
            currPath.pop_back();
        }
    }

    void bfs(string &beginWord, string &endWord,
             unordered_set<string> &wordList) {
        queue<string> q;
        q.push(beginWord);

        // remove the root word which is the first layer
        if (wordList.count(beginWord)) {
            wordList.erase(beginWord);
        }

        unordered_set<string> isEnqueued;
        isEnqueued.insert(beginWord);

        while (!q.empty()) {
            // visited will store the words of current layer
            vector<string> visited;

            for (int i = q.size() - 1; i >= 0; i--) {
                string currWord = q.front();
                q.pop();

                // findNeighbors will have the adjacent words of the currWord
                vector<string> neighbors = findNeighbors(currWord, wordList);
                for (auto word : neighbors) {
                    visited.push_back(word);
                    // add the edge from word to currWord in the list
                    adjList[word].push_back(currWord);

                    // de-duplicate
                    if (!isEnqueued.count(word)) {
                        q.push(word);
                        isEnqueued.insert(word);
                    }
                }
            }
            // removing the words of the previous layer
            for (int i = 0; i < visited.size(); i++) {
                if (wordList.count(visited[i])) {
                    wordList.erase(visited[i]);
                }
            }
        }
    }

    vector<vector<string>> findLadders(string beginWord, string endWord,
                                       vector<string> &wordList) {
        // copying the words into the set for efficient deletion in BFS
        unordered_set<string> copiedWordList(wordList.begin(), wordList.end());
        // build the DAG using BFS
        bfs(beginWord, endWord, copiedWordList);

        // every path will start from the endWord
        currPath = {endWord};
        // traverse the DAG to find all the paths between endWord and beginWord
        // 这里是从endWord开始来找beginWord，而且前面的adjList也是倒着来建的。
        backtrack(endWord, beginWord);

        return shortestPaths;
    }
};
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