<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:
    int eq(string &s,string&t)
    {  int count=0;
        for(int i=0;i<s.size();i++)
        {   
            if(s[i]!=t[i])count++;
            if(count>1)return 0;

        }

        return 1;
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList)
    { int br=1;
           for(auto it:wordList)
           {
            if(it==beginWord)br=0;
           }
   
   if(br)wordList.push_back(beginWord);

        vector<pair<string,int>>vt;
  int start,end;
        int i=0,n=wordList.size();
        for(auto it:wordList)
        {       if(it==beginWord)start=i;
                if(it==endWord)end=i;
                vt.push_back({it,i++});
        }

        vector<vector<int>>list(n);

        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
               if(eq(wordList[i],wordList[j]))
               {
                list[i].push_back(j);
                list[j].push_back(i);
               }
            }

        }



      int visited[5005]={};

      queue<pair<int,int>>qu;

      qu.push({start,0});
      visited[start]=1;
      while(!qu.empty())
      {
        auto it=qu.front();
        qu.pop();
        if(it.first==end)return it.second+1;
         
         for(auto itt:list[it.first])
         {
            if(visited[itt]==0)
            {   visited[itt]=1;
                qu.push({itt,it.second+1});
            }
         }


      }

        

return 0;
    }
};
25875