<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 {
private:
    vector<string> transform(string& word, unordered_set<string>& wordSet, unordered_set<string>& seen) {
        vector<string> nextWords;

        for (int i = 0; i < word.length(); i++) {
            string fuzzWord = word;
            for (char ch = 'a'; ch <= 'z'; ch++) {
                if (word[i] == ch)
                    continue; 
                
                fuzzWord[i] = ch; 

                if (wordSet.find(fuzzWord) != wordSet.end() && seen.find(fuzzWord) == seen.end()) {
                    nextWords.push_back(fuzzWord);
                    seen.insert(fuzzWord);
                    // wordSet.erase(fuzzWord);
                }
            }
        }
        return nextWords;
    }

public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> wordSet (wordList.begin(), wordList.end());
        unordered_set<string> seen;

        int wordLen = beginWord.length();

        if (wordSet.find(endWord) == wordSet.end())
            return 0;

        queue<string> todo;
        todo.push(beginWord);
        seen.insert(beginWord);

        int length = 1;
        while (!todo.empty()) {
            int n_todo = todo.size();
            for (int i = 0; i < n_todo; i++) {
                string currWord = todo.front();
                todo.pop();

                // cout << currWord << endl;
                if (currWord == endWord)
                    return length;

                vector<string> nextWords = transform(currWord, wordSet, seen);
                for (string nextWord : nextWords)
                    todo.push(nextWord);

            }
            length++;
        }

        return 0;
    }
};
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