<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:
    unordered_map<string,vector<string>> mp;
    unordered_map<string,bool> visited;
    bool diff(string s1 , string s2){
        int i = 0;
        int res = 0;
        while(i < s1.size()){
            if(s1[i] != s2[i]){
                res++;
            }
            if(res > 1) return false;
            i++;
        }
        return true;
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        vector<string> simil;
        for(int j = 0 ; j < wordList.size() ; j++){
            if(diff(beginWord , wordList[j])){
                simil.push_back(wordList[j]);
                mp[wordList[j]].push_back(beginWord);
            }
            visited[wordList[j]] = false;
        }
        if(simil.empty()){
            return 0;
        }
        mp[beginWord] = simil;
        for(int i = 0 ; i < wordList.size() - 1 ; i++){
            string curr = wordList[i];
            for(int j = i + 1 ; j < wordList.size() ; j++){
                if(diff(curr , wordList[j])){
                    mp[curr].push_back(wordList[j]);
                    mp[wordList[j]].push_back(curr);
                }
            }
        }
        if(mp.find(endWord) == mp.end()){
            return 0;
        }
        // for(auto itr : mp){
        //     cout<<itr.first<<" --> ";
        //     for(string i : itr.second){
        //         cout<<i<<" ";
        //     }
        //     cout<<endl;
        // }
        int level = 0;
        deque<string> dq;
        dq.push_back(beginWord);
        dq.push_back("");
        while(!dq.empty()){
            while(dq.front()!= ""){
                string curr = dq.front();
                dq.pop_front();
                visited[curr] = true;
                for(string itr : mp[curr]){
                    if(!visited[itr]){
                        dq.push_back(itr);
                    }
                }
            }
            dq.pop_front();
            if(!dq.empty()){
                dq.push_back("");
            }
            level++;
            if(visited[endWord]){
                break;
            }
        }
        return (visited[endWord]) ? level : 0;
    }
};
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