<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, vector<string>> adjList; 
        unordered_map<string, bool> visited; 

        bool transPos = false;
        int L = beginWord.length();
        for (auto& word : wordList) {
            if (word == endWord) {
                transPos = true; 
            }

            for (int i = 0; i < L; i++) {
                string w = word.substr(0, i) + "*" + word.substr(i+1, L); 
                adjList[w].push_back(word);
            } 
        }

        if (!transPos) return 0; 
        
        queue<pair<string, int>> q; 
        q.push({beginWord, 0}); 
        while (!q.empty()) {
            auto e = q.front(); q.pop();
            visited[e.first] = true;
            if (e.first == endWord) return e.second+1; 
            for (int i = 0; i < L; i++) {
                string w = e.first.substr(0, i) + "*" + e.first.substr(i+1, L); 
                for (auto& n : adjList[w]) {
                    if (visited[n]) continue;
                    q.push({n,e.second + 1}); 
                }
            }

        } 


        // for (auto& t : adjList) {
        //     cout << t.first << ": "; 
        //     for (auto& e : t.second) {
        //         cout << e << " "; 
        //     } cout << endl; 
        // }
        return 0;
    }
};



/* 
        lot -- log -- cog 
       /  \      |    /
      /    \     |   /
    hot -- dot -- dog 
     |
     |   
    hit
*/
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