<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:
    std::unordered_set<std::string> setified_wordlist {};

public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        // Shortest path - BFS (shortest transformation)
        // Let n = wordList.size()
        // (1) V(G) = all words in wordList, E(G) is all possible linkages between words (n^2)
        // C = beginWord.length()
        // Then runtime is n * (C * n^2) = C * n^3

        // (2) V(G) = all words in wordList, but use a "partially unmaterialized" dict
        // Construction of "globbed dict": n * C^2
        // BUT, the size of a globbed dict entry is at most 26...
        // BFS runtime: n * (C^2 * min(n-1, 26)) = n C^2

        for (auto word : wordList) {
            setified_wordlist.insert(word);
        }

        if (!setified_wordlist.contains(endWord)) {
            return 0;
        }

        std::unordered_map<std::string, std::unordered_set<std::string>> glob_dict {};
        char temp;
        std::string str_save;
        for (auto str : wordList) {
            str_save = str;
            for (int i = 0; i < str.length(); i++) {
                temp = str[i];
                str[i] = '*';
                if (!glob_dict.contains(str)) {
                    glob_dict[str] = std::unordered_set<std::string> {str_save};
                } else {
                    glob_dict[str].insert(str_save);
                }
                str[i] = temp;
            }
        }

        // std::unordered_map<std::string, std::unordered_set<std::string>> glob_dict {};
        // TODO: see what's in `glob_dict`
        // for (auto pair : glob_dict) {
        //     std::cout << pair.first << ' ' << pair.second.size() << '\n';
        // }


        // BFS
        std::deque<std::string> Q {};
        std::unordered_map<std::string, int> dist {};
        // cold start init!!
        Q.push_back(beginWord);
        dist[beginWord] = 1; // we're counting # of vertices, I guess

        std::string vertex;
        std::string cand;
        while (!Q.empty()) {
            vertex = Q.front();
            Q.pop_front();

            for (int i = 0; i < vertex.length(); i++) {
                cand = vertex;
                cand[i] = '*';
                for (auto outNeighbor : glob_dict[cand]) {
                    if (!dist.contains(outNeighbor)) {
                        dist[outNeighbor] = dist[vertex] + 1;
                        // Add to the queue, dummy
                        Q.push_back(outNeighbor);
                    }
                }
            }
        }


        // See what's in dist
        // for (auto pair : dist) {
        //     std::cout << pair.first << ' ' << pair.second << '\n';
        // }


        return (!dist.contains(endWord)) ? 0 : dist[endWord];
    }
};
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