<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:
    
    bool canBeNextNode (string a, string b) {
        int count = 0;
        for (int i = 0; i < a.size() && count <= 1; i++)
            if (a[i] != b[i])
                count++;
        
        return count == 1;
    }

    unordered_map<string, vector<string>> createGraph (string beginWord, vector<string> &wordList1) {
        
        vector<string> wordList(wordList1);
        wordList.push_back(beginWord);
        
        unordered_map<string, vector<string>> resultMap;
        for (auto word : wordList) {
            vector<string> v;
            resultMap[word] = v;
        }
        
        for (int i = 0; i < wordList.size(); i++) {
            for (int j = i + 1; j < wordList.size(); j++) {
                if (canBeNextNode(wordList[i], wordList[j])) {
                    resultMap[wordList[i]].push_back(wordList[j]);
                    resultMap[wordList[j]].push_back(wordList[i]);
                }
            }
        }

        return resultMap;
    }
        
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_map<string, vector<string>> wordGraph = createGraph(beginWord, wordList);
        unordered_map<string, bool> visited;
        
        queue<string> q, qClone, temp;
        q.push(beginWord);
        
        int pathLength = 1;
        
        while (q.size()) {
            
            while(q.size()) {
                string currWord = q.front(); q.pop();
                
                if (currWord == endWord)
                    return pathLength;
                
                if (!visited[currWord])
                    for (auto nextPossible : wordGraph[currWord])
                        qClone.push(nextPossible);
                
                visited[currWord] = true;
            }
            
            temp = q;
            q = qClone;
            qClone = temp;
            
            pathLength++;
        }
        
        return 0;
    }
    
};
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