<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 find(string ref,vector<string>&word,map<string,vector<string>>&adj,int size){
        int x=word.size();
        for(int i=0;i<x;i++){
            int cnt=0,j=0;
            string chk=word[i];
            while(j<size){
                if(ref[j]!=chk[j])cnt++;
                j++;
            }
            if(cnt==1){
                adj[ref].push_back(chk);
                adj[chk].push_back(ref);
            }
        }

    }
    int bfs(string src,map<string,vector<string>>adj,set<string>&s,string e){
        s.insert(src);
        queue<pair<string,int>>q;
        q.push({src,1});
        while(!q.empty()){
            string word=q.front().first;
            int cnt=q.front().second;
            q.pop();
            for(auto it:adj[word]){
                if(s.find(it)==s.end()){
                    if(it==e)return cnt+1;
                    else{
                        q.push({it,cnt+1});
                        s.insert(it);
                    }
                }
            }

        }
        return 0;
    }
    int ladderLength(string b, string e, vector<string>& word) {
        bool check=0;
        for(auto it:word){
            if(it==e)check=1;
        }
        if(check==0)return 0;
        map<string,vector<string>>adj;
        int x=b.size();
        find(b,word,adj,x);
        int size=word.size();
        for(int i=0;i<size;i++){
            if(word[i]!=b){
                find(word[i],word,adj,x);
            }
        }
        set<string>vis;
        int ans=bfs(b,adj,vis,e);
        return ans;

    }
};
58375