<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>

struct Node
{
    bool is_visited_ = false;
    const string * word_ptr_;
    vector<Node*> neighbors_;

    Node(const string & word) : word_ptr_(&word) {}

    bool is_visited() const
    {
        return is_visited_;
    }

    void set_visited()
    {
        is_visited_ = true;
    }

    void connect_neighbor(Node * n)
    {
        int cnt = 0;
        for (int i = 0; i < word_ptr_->size(); i++)
        {
            cnt += word_ptr_->at(i) != n->word_ptr_->at(i);
            if (cnt == 2)
            {
                break;
            }
        }

        if (cnt == 1)
        {
            neighbors_.push_back(n);
            n->neighbors_.push_back(this);
        }
    }
};

class Solution {
public:
    int ladderLength(string s1, string s2, vector<string>& words)
    {
        auto vit2 = find(words.begin(), words.end(), s2);
        if (vit2 == words.end())
        {
            return 0;
        }

        vit2->swap(words.front());    // front --> s2

        auto vit1 = find(words.begin(), words.end(), s1);
        if (vit1 == words.end())
        {
            words.push_back(s1);      // back  --> s1
        }
        else
        {
            vit1->swap(words.back()); // back  --> s1
        }
        

        vector<Node*> nodes;
        nodes.reserve(words.size());
        for (const auto & word : words)
        {
            auto ni = new Node(word);
            for (auto nj : nodes)
            {
                ni->connect_neighbor(nj);
            }
            nodes.push_back(ni);
        }

        Node* n1 = nodes.back();
        Node* n2 = nodes.front();
        n1->set_visited();

        queue<Node*> q;
        q.push(n1);
        int cnt = 0;
        while (!q.empty())
        {
            cnt++;
            for (int i = 0, iend = q.size(); i < iend; i++)
            {
                Node * ni = q.front();
                if (ni == n2)
                {
                    return cnt;
                }

                q.pop();
                for (auto nj : ni->neighbors_)
                {
                    if (nj->is_visited())
                    {
                        continue;
                    }

                    nj->set_visited();
                    q.push(nj);
                }
            }
        }
        return 0;
    }
};
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