<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) {
        // build transformation graph (including beginWord)
        if (find(wordList.begin(), wordList.end(), beginWord) == wordList.end()) {
            wordList.push_back(beginWord);
        }
        map<string, vector<string>> graph;
        map<string, int> idx;
        for(int i = 0; i < wordList.size(); i++) {
            string x = wordList[i];
            idx[x] = i;
            for(int j = i+1; j < wordList.size(); j++) {
                string y = wordList[j];
                if (x == y) continue;
                if (x.size() != y.size()) continue;
                int common = 0;
                for(int i = 0; i < x.size(); i++) {
                    if (x[i] == y[i]) common++;
                }
                if (common == x.size()-1) {
                    graph[x].push_back(y);
                    graph[y].push_back(x);
                }
            }
        }
        // for(auto p : idx) cout << p.first << " " << p.second << endl;
        // for(string x : wordList) {
        //     cout << x << ": ";
        //     for (string y : graph[x]) {
        //         cout << y << " ";
        //     }
        //     cout << endl;
        // }
        // shortest path from beginWord to endWord (bfs)
        queue<string> q;
        set<int> seen;
        q.push(beginWord);
        int t = 0;
        while(!q.empty() && t <= wordList.size()) {
            int o = q.size();
            t++;
            while(o--) {
                string x = q.front(); q.pop();
                if (x == endWord) return t;
                seen.insert(idx[x]);
                for(string y : graph[x]) {
                    if (seen.count(idx[y])) {
                        continue;
                    }
                    q.push(y);
                }
            }
        }
        return 0;
    }
};
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