<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> s1;
        for(string&x:wordList) s1.insert(x);
        unordered_map<string, vector<string>> m1;
        unordered_map<string, int> dis;
        for (const string& word : wordList) dis[word] = INT_MAX;
        dis[beginWord] = 0;
        queue<string> q1;
        q1.push(beginWord);
        while(!q1.empty()){
            string temp=q1.front();
            q1.pop();
            for(int j=0;j<temp.length();j++){
                for(int k=0;k<26;k++){
                    char rep='a'+k;
                    string temp1=temp;
                    temp1[j]=rep;
                    if(s1.find(temp1)!=s1.end()){
                        m1[temp].push_back(temp1);
                        m1[temp1].push_back(temp);
                        q1.push(temp1);
                        s1.erase(temp1);
                    }
                }
            }
        }
        priority_queue<pair<int, string>, vector<pair<int, string>>, greater<pair<int, string>>> pq;
        pq.push({0, beginWord});
        while (!pq.empty()) {
            auto [currentDist, word] = pq.top();
            pq.pop();

            if (word == endWord) {
                return currentDist + 1; 
            }

            for (const string& neighbor : m1[word]) {
                int newDist = currentDist + 1;
                if (newDist < dis[neighbor]) {
                    dis[neighbor] = newDist;
                    pq.push({newDist, neighbor});
                }
            }
        }
        return 0; 
    }
};
30425