<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>

ostream& operator<<(ostream& os, unordered_map<string, vector<int>>& m){
    for(const pair<string,vector<int>>& p : m){
        os << endl;
        os << p.first << " : " << endl;
        for(int n : p.second){
            os << n << "  " ;
        }
    }
    return os;
} 

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        // create a adjacent graph. connect two nodes only if they are diifer by exact one letter.
        // do BSF (breadth search first) if reach the endWord successfully, return the number of edge + 1.
        // if beginWord does not exist in wordList, add it to the end of wordList.
        
        if(find(wordList.begin(), wordList.end(), endWord) >= wordList.end()){
            return 0;
        }
        int end_index = 0;
        for(int i=0;i<wordList.size();++i){
            if(wordList[i]==endWord){
                end_index = i;
                break;
            }
        }

        unordered_map<string, vector<int>> my_map; 
        for(int i=0; i<wordList.size();++i){
            string word = wordList[i];
            for(int j=0;j<word.size();++j){
                string tmp = word;
                tmp[j] = '*';
                my_map[tmp].push_back(i);
            }
        }
        
        vector<int> transformed(wordList.size(), -1);
        queue<int> my_queue;
        for(int i=0;i<beginWord.size();++i){
            string tmp = beginWord;
            tmp[i] = '*';
            for(int index : my_map[tmp]){
                my_queue.push(index);
                transformed[index]=1;
            }
        }
        while(!my_queue.empty()){
            int index = my_queue.front();
            my_queue.pop();
            for(int i=0;i<endWord.size();++i){
                string tmp = wordList[index];
                tmp[i] = '*';
                for(int next : my_map[tmp]){
                    if(transformed[next] == -1){
                        transformed[next] = 1 + transformed[index];
                        if(next == end_index){
                            break;
                        }
                        my_queue.push(next);
                    }
                }
            }
        }
        return transformed[end_index]+1;
    }
};
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