<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) {
        vector<string> palavras;
        map<string, int> mapa;
        int n = 1;

        palavras.push_back(beginWord);
        mapa[beginWord] = 0;

        for (string &s : wordList){
            if (mapa.find(s) == mapa.end()){
                palavras.push_back(s);
                mapa[s] = n;
                n++;
            }
        }

        if (mapa.find(endWord) == mapa.end()) return 0;
        vector<vector<int>> grafo(n);

        for (int i=0; i<n; ++i){
            for (int j=i+1; j<n; ++j){
                string a = palavras[i];
                string b = palavras[j];

                if (a.size() != b.size()) continue;
                int dif = 0;

                for (int k=0; k<a.size(); ++k){
                    if (a[k] != b[k]) dif++;
                    if (dif > 1) break;
                }

                if (dif == 1){
                    grafo[i].push_back(j);
                    grafo[j].push_back(i);
                }
            }
        }

        int start = mapa[beginWord];
        int end = mapa[endWord];

        queue<int> fila;
        fila.push(start);
        vector<int> dist(n, -1);
        dist[start] = 1;
        int ans = 0;

        while (!fila.empty()){
            int u = fila.front();
            fila.pop();

            if (u == end){
                ans = dist[u];
                break;
            }

            for (int v : grafo[u]){
                if (dist[v] == -1){
                    dist[v] = dist[u] + 1;
                    fila.push(v);
                }
            }
        }

        return ans;
    }
};
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