<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 {

    int bfs(vector<vector<int>>& graph,int startNode,string beginWord,string endWord,vector<string>& wordList){
        queue<pair<int,int>> q;
        vector<bool> vis(graph.size());
        q.push({startNode,1});
        while(!q.empty()){
            pair<int,int> fr=q.front();
            q.pop();
            if(vis[fr.first])continue;
            vis[fr.first]=true;
            string str=fr.first==graph.size()-1?beginWord:wordList[fr.first];
            if(str==endWord)return fr.second;
            for(int i=0;i<graph[fr.first].size();i++){
                if(!vis[graph[fr.first][i]])q.push({graph[fr.first][i],fr.second+1});
            }
        }
        return 0;
    }


public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int n=wordList.size();
        vector<vector<int>> graph(n+1);
        map<string,int> m;
        for(int i=0;i<=n;i++){
            string str=i==n?beginWord:wordList[i];
            m[str]=i;
        }   
        for(int i=0;i<=n;i++){
            string str=i==n?beginWord:wordList[i];            
            for(int j=0;j<str.size();j++){
                char old=str[j];
                for(char c='a';c<='z';c++){
                    if(c==old)continue;
                    str[j]=c;
                    if(m.find(str)!=m.end()){
                        graph[i].push_back(m[str]);
                        graph[m[str]].push_back(i);
                    }
                }
                str[j]=old;
            }
        }
        return bfs(graph,n,beginWord,endWord,wordList);
    }
};
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