<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:
    static constexpr int N = 5000;
    string* dict[N];
    string* path[N];   
    int word_len = 0;

    bool candidate(string* curr, string* word) {
        int count = 0;
        for (int i = 0; i < word_len; i++) {
            if ((*curr)[i] != (*word)[i]) {
                count++;
            }
        }
        return count == 1;
    }
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        word_len = beginWord.length();
        int dict_len = 0;
        for (auto& w : wordList) {
            dict[dict_len++] = &w;
        }
        int path_len = 0;        
        path[path_len++] = &beginWord;
        bool added = true;
        int start_len = 0;
        int steps = 1;
        while (added) {
            steps++;
            added = false;
            //std::cout << "step " << count << "\n";            
            int curr_len = path_len;
            for (int i = start_len; i < curr_len; i++) {
                string* curr = path[i];
                for (int j = 0; j < dict_len; j++) {
                    string* word = dict[j];
                    if (candidate(curr, word)) {
                        if (*word == endWord) {
                            return steps;
                        }
                        //std::cout << "added " << w << "\n";
                        path[path_len++] = word;
                        added = true;
                        dict[j--] = dict[--dict_len];                  
                    }
                }              
            }
            start_len = curr_len;
        }
        return 0;
    }
};
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