<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_map<string, int> mp;
        int n = wordList.size();
        string str;
        char c, prev;
        for(int i = 0; i < n; i++) {
            mp[wordList[i]] = i;
        }
        if(mp.find(beginWord) == mp.end()) {
            wordList.push_back(beginWord);
            mp[beginWord] = n;
            n++;
        }
        vector<vector<int>> graph(n);
        for(int k = 0; k < wordList.size(); k++) {
            str = wordList[k];
            for(int i = 0; i < 26; i++) {
                c = char('a' + i);
                for(int j = 0; j < wordList[k].size(); j++) {
                    prev = wordList[k][j];
                    str[j] = c;
                    if(prev != c && mp.find(str) != mp.end()) {
                        graph[k].push_back(mp[str]);
                        graph[mp[str]].push_back(k);
                    }
                    str[j] = prev;
                }
            }
        }
        queue<pair<int, int>> q;
        q.push({mp[beginWord], 1});
        vector<int> vis(n, 0);
        pair<int, int> curr;
        int ind;
        while(!q.empty()) {
            curr = q.front();
            q.pop();
            ind = curr.first;
            if(wordList[ind] == endWord) {
                return curr.second;
            }
            if(vis[curr.first] == 1) {
                continue;
            }
            vis[curr.first] = 1;
            for(int i = 0; i < graph[ind].size(); i++) {
                if(vis[graph[ind][i]] == 0) {
                    q.push({graph[ind][i], curr.second + 1});
                }
            }
        }
        return 0;
    }
};
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