<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>the <strong>number of words</strong> in the <strong>shortest transformation sequence</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or </em><code>0</code><em> if no such sequence exists.</em></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> 5
<strong>Explanation:</strong> One shortest transformation sequence is &quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; cog&quot;, which is 5 words long.
</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> 0
<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;= 10</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 5000</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>
</ul>

class Solution {
public:
    
    void fill_with_neighbour(string &x, unordered_map<string, vector<string> > &mp){
        string s = x; 
        int n = x.size();
        mp[x] = {};
        for(int j = 0; j < n; j ++){
            for ( int i = 0 ; i < 26 ; i ++){
                s[j] = char('a' + i);
                if (s[j] != x[j] && mp.find(s) != mp.end()){
                    mp[x].push_back(s);
                    mp[s].push_back(x);
                }  
            }
            s[j] = x[j];
        }
        
    }
    
    
    int ladderLength(string str, string str2, vector<string>& ar) {
        
        unordered_map<string, vector<string> > mp;
        mp[str] = {};
        int res = 0;
        
        for(auto x : ar){ 
            fill_with_neighbour(x, mp);
        }
        
        if (mp.find(str2) == mp.end())
            return res;
        
        unordered_map<string, int > vis;
        queue<string > q; 
        q.push(str);

        
        while(q.size()){
            res ++; 
            int len = q.size(); 
            while(len --){
                string s = q.front(); q.pop(); 
                vis[s] = 1;
                if (s == str2)
                    return res; 
                for(auto x : mp[s]){
                    if (!vis[x])
                        q.push(x);
                }
            }
        }
        
        return 0;
        
        
    }
};
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