<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) {

        unordered_set<string> uset;
        uset.insert(beginWord);
        for(string word: wordList)
            uset.insert(word);
        
        unordered_map<string,vector<string>> umap;

        for(int i = 0 ; i < beginWord.size(); i++){
            string tmp = beginWord;
            for(char c='a'; c<='z'; c++){
                tmp[i]=c;
                if(uset.find(tmp)!=uset.end()){
                    umap[beginWord].push_back(tmp);
                }
            }
        }
        for(string& adyacent:wordList){
            for(int i = 0 ; i < beginWord.size(); i++){
                string tmp = adyacent;
                for(char c='a'; c<='z'; c++){
                    tmp[i]=c;
                    if(uset.find(tmp)!=uset.end()){
                        umap[adyacent].push_back(tmp);
                    }
                }
            }
        }

        // for(auto it:umap){
        //     cout<<it.first<<" = ";
        //     for(string s:it.second){
        //         cout<<s<<" ";
        //     }
        //     cout<<endl;
        // }

        deque<string> dq;
        dq.push_back(beginWord);
        int iteration = 0;

        while(dq.size() != 0){
            int dqSize = dq.size();

            for(int i = 0 ; i < dqSize; i++){
                string currWord = dq.front(); //hit
                dq.pop_front();
                if(currWord == endWord){
                    return iteration+1;
                }
                for(string& nextWord:umap[currWord]){
                    dq.push_back(nextWord);
                }
                umap[currWord] = {};
            }
            iteration++;
        }

        return 0;
        
    }
};
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