<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:
    unordered_map<string, vector<string>> graph;

    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int m = wordList.size();
        int n = wordList[0].size();
        bool endWordFound = false;

        int offBy = 0;

        for (int i = 0; i < m; i++) {
            if (wordList[i] == endWord) endWordFound = true;
            for (int j = 0; j < n; j++) {
                if (beginWord[j] != wordList[i][j]) offBy++;
            }
            if (offBy == 1) {
                graph[beginWord].push_back(wordList[i]);
                graph[wordList[i]].push_back(beginWord);
            }
            offBy = 0;
        }

        if (!endWordFound) return 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < m; j++) {
                if (i == j) continue;

                for (int k = 0; k < n; k++) {
                    if (wordList[i][k] != wordList[j][k]) {
                        offBy++;
                    }
                }

                if (offBy == 1) {
                    graph[wordList[i]].push_back(wordList[j]);
                    graph[wordList[j]].push_back(wordList[i]);
                }

                offBy = 0;
            }
        }

        unordered_set<string> seen({beginWord});
        queue<string> que({beginWord});
        int depth = 1;

        while (!que.empty()) {
            int width = que.size();

            for (int i = 0; i < width; i++) {
                string curr = que.front();
                que.pop();

                if (curr == endWord) return depth;

                for (auto word : graph[curr]) {
                    if (!seen.count(word)) {
                        seen.insert(word);
                        que.push(word);
                    }
                }
            }

            depth++;
        }

        return 0;
    }
};
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