<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 cmpStrng(string s1, string s2) {
    int cnt = 0;
    if (s1.size() != s2.size()) return -1;
    
    for (int i = 0; i < s1.size(); i++) {
        if (s1[i] != s2[i]) cnt++;
    }
    return cnt;
}

int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
    if (beginWord.size() != endWord.size()) return 0;  

    wordList.insert(wordList.begin(), beginWord);
    
    int n = wordList.size();  
    unordered_map<int, vector<int>> adj;
    int endWordIdx=-1;
    for (int i = 0; i < n; i++) {
            if(wordList[i]==endWord)endWordIdx=i;
        for (int j = i + 1; j < n; j++) {
            if (cmpStrng(wordList[i], wordList[j]) == 1) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }
    
   

    if(endWordIdx==-1)return 0;
    vector<int>dist(n,1e9);
    queue<int>q;
    q.push(0);
    dist[0]=0;
    while(!q.empty()){
        int front=q.front();
        q.pop();
        for(auto neigh:adj[front]){
            if(dist[neigh]>1+dist[front]){
                dist[neigh]=dist[front]+1;
                q.push(neigh);
            }
        }
    }


    return dist[endWordIdx]==1e9?0:dist[endWordIdx]+1;  
}

};
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