<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 Node{
    public:
        string val;
        vector<Node*> list;
        Node(string _val){
            val=_val;
        }
};

class Solution {
public:
    int ladderLength(string startGene, string endGene, vector<string>& bank) {
        map<string, Node*> graph;
        map<string, bool> visited;
        bool hasStart = false;
        for(int i=0;i<bank.size();i++){
            graph[bank[i]]=new Node(bank[i]);
            if(bank[i]==startGene) hasStart=true;
        }
        if(!hasStart){
            graph[startGene] = new Node(startGene);
            bank.push_back(startGene);
        }
        for(int i=0;i<bank.size();i++){
            for(int j=i+1;j<bank.size();j++){
                int count=0;
                for(int k=0;k<bank[j].size();k++){
                    if(bank[i][k]!=bank[j][k]) count++;
                }
                if(count==1){
                    // cout<<bank[i]<<"->"<<bank[j]<<endl;
                    graph[bank[i]]->list.push_back(graph[bank[j]]);
                    graph[bank[j]]->list.push_back(graph[bank[i]]);
                }
            }
        }
        queue<Node*> q;
        q.push(graph[startGene]);
        visited[startGene] = true;
        q.push(NULL);
        int ans=1;
        while(!q.empty()){
            Node* curr=q.front();
            q.pop();
            if(curr==NULL){
                ans++;
                if(q.empty()) return 0;
                q.push(NULL);
            }else{
                if(curr->val==endGene) return ans;
                for(auto a:curr->list){
                    if(!visited[a->val]){
                        visited[a->val] = true;
                        q.push(a);
                    }
                }
            }
        }
        return ans;
    }
};
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