<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 ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        if(find(wordList.begin(),wordList.end(),endWord)==wordList.end()){return 0;}
        unordered_map<string,vector<string>> dist1;
        wordList.push_back(beginWord);
        for(int i=0;i<wordList.size();i++){
            for(int j=i+1;j<wordList.size();j++){
                if(wordList[i].size()!=wordList[j].size()){continue;}
                int dist=0;
                for(int k=0;k<wordList[i].size();k++){
                    if(wordList[i][k]!=wordList[j][k]){dist++;}
                }
                if(dist==1){dist1[wordList[i]].push_back(wordList[j]);
                    dist1[wordList[j]].push_back(wordList[i]);
                }
            }
        }
        wordList.pop_back();
        // for(auto a:wordList){
        //     // if(wordList[i].size()!=wordList[j].size()){continue;}
        //         int dist=0;
        //         for(int k=0;k<beginWord.size();k++){
        //             if(beginWord[k]!=a[k]){dist++;}
        //         }
        //         if(dist==1){dist1[beginWord].push_back(a);
        //             dist1[a].push_back(beginWord);
        //         }
        // }
        queue<string> bfs;
        set<string> visit;
        bfs.push(beginWord);
        int len=0;
        while(!bfs.empty()){
            int siz=bfs.size();
            len++;
            while(siz){
                string topo=bfs.front();
                visit.insert(topo);
                if(topo==endWord){
                    return len;
                }
                bfs.pop();
                siz--;
                for(auto a:dist1[topo]){
                    if(visit.find(a)==visit.end()){
                    bfs.push(a);}
                }
            }
        }
        return 0;

    }
};
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