<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:
    bool fn(string s1,string s2){   
        int count=0;
        for(int i=0;i<s1.size();i++){
            if(s1[i]!=s2[i])
                count++;
        }
        if(count==1)
            return true;
        else return false;
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
       unordered_map<string,bool> visited;
       unordered_set<string> book;
       int n=beginWord.size();
    //    for(int i=0;i<wordList.size();i++){
    //     book.insert(wordList[i]);
    //    }
       unordered_map<string,list<string>> adj;
       wordList.push_back(beginWord);
       for(int i=0;i<wordList.size();i++){
            for(int j=i+1;j<wordList.size();j++){
                if(fn(wordList[i],wordList[j])){
                    adj[wordList[i]].push_back(wordList[j]);
                    adj[wordList[j]].push_back(wordList[i]);
                }
            }
       }
//         vector<char> characters={
//     'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'
// };
       queue<string>q;
       q.push(beginWord);
       q.push("0");
       visited[beginWord]=true;
       int count=0;
    
       while(!q.empty()){
        string next=q.front();
        cout<<next<<" ";
        q.pop();
        if(next=="0"){
            if(q.empty()){
                return 0;
            }
            count++;
            q.push("0");
            continue;
       }
        for(auto s :adj[next]){
            if(s==endWord)
                return count+2;
            if(!visited[s]){
                visited[s]=true;
                q.push(s);
            }
        }
        // for(int i=0;i<n;i++){
        //      string stri=next;
        //     for(auto ch : characters){
        //         stri[i]=ch;
                
        //         if(book.count(stri)&& !visited[stri]){
        //             if(stri==endWord)
        //                 return count+2;
        //             visited[stri]=true;
        //             q.push(stri);
        //         }
        //     }
        // }
       }

       return count+1;
        
    }


};
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