<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> mp;
    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 steps=mp[word];
        int size=word.size();

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

            for(char ch='a';ch<='z';ch++)
            {
                word[i]=ch;

                if(mp.find(word)!=mp.end() && mp[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) {

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

        int size=beginWord.size();
        b=beginWord;

        unordered_set<string> st(wordList.begin(), wordList.end());
        st.erase(beginWord);

        mp[beginWord]=1;

        while(!q.empty())
        {
            string word=q.front();
            int steps=mp[word];
            q.pop();

            if(word==endWord)
            {
                break;
            }

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

                for(char ch='a';ch<='z';ch++)
                {
                    word[i]=ch;

                    if(st.find(word)!=st.end())
                    {
                        q.push({word});
                        st.erase(word);
                        mp[word]=steps+1;
                    }
                }

                word[i]=original;
            }
        }
        
        if(mp.find(endWord)!=mp.end())
        {
            vector<string> seq;
            seq.push_back(endWord);
            dfs(endWord, seq);
        }

        return ans;
    }
};
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