<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 {
    unordered_map<string, int> mpp;
    vector<vector<string>> ans;
    string b;

private:
    void dfs(string word, vector<string>& seq) {
        if (word == b) {
            reverse(seq.begin(), seq.end());
            ans.push_back(seq);

            reverse(seq.begin(), seq.end());
            return;
        }
        int sz = word.size();
        int steps = mpp[word];
        for (int i = 0; i < sz; i++) {
            char original = word[i];
            for (char ch = 'a'; ch <= 'z'; ch++) {
                word[i] = ch;
                if (mpp.find(word) != mpp.end() && mpp[word] + 1 == steps) {
                    seq.push_back(word);
                    dfs(word, seq);
                    seq.pop_back();
                }
            }
            word[i] = original;
        }
    }

public:
    vector<vector<string>> findLadders(string beginWord, string endWord,
                                       vector<string>& wordList) {
        unordered_set<string> st(wordList.begin(), wordList.end());

        queue<string> q;
        b = beginWord;
        q.push({beginWord});

        mpp[beginWord] = 1;
        int sz = beginWord.size();
        st.erase(beginWord);
        while (!q.empty()) {

            string word = q.front();
            int steps = mpp[word];
            q.pop();

            if (word == endWord)
                break;

            for (int i = 0; i < sz; i++) {
                char original = word[i];

                for (char ch = 'a'; ch <= 'z'; ch++) {

                    word[i] = ch;
                    if (st.count(word)) {
                        q.push(word);
                        st.erase(word);

                        mpp[word] = steps + 1;
                    }
                }
                word[i] = original;
            }
        }

        if (mpp.find(endWord) != mpp.end()) {
            vector<string> seq;
            seq.push_back(endWord);
            dfs(endWord, seq);
        }
        return ans;
    }
};
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