<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) {
        AddWordToWildCards(beginWord);
        AddWordToWildCards(endWord);
        bool endWordInWordList = false;
        for (auto& w : wordList) {
            if (w == endWord) {
                endWordInWordList = true;
            }
            AddWordToWildCards(w);
        }
        if (!endWordInWordList) {
            return 0;
        }

        BuildGraph(beginWord, endWord, wordList);

        // BFS
        queue<int> q;
        vector<bool> visited(m_ids.size());

        q.push(GetId(beginWord));
        int targetId = GetId(endWord);
        int res = 1;

        while (!q.empty()) {
            int size = q.size();
            for (int i = 0;i < size;i++) {
                int cur = q.front();
                q.pop();
                if (cur == targetId) {
                    return res;
                }

                for (int neighbor : m_neighbors[cur]) {
                    if (!visited[neighbor]) {
                        q.push(neighbor);
                        visited[neighbor] = true;
                    }
                }
            }

            res++;
        }
        return 0;
    }

private:
    int GetId(const string& word) {
        return m_ids.try_emplace(word, m_ids.size()).first->second;
    }

    void AddWordToWildCards(string& word) {
        int wordId = GetId(word);
        for (size_t i = 0;i < word.size();i++) {
            char backup = word[i];
            word[i] = '*';
            m_wildcardToWords[GetId(word)].push_back(wordId); // This wild card can connect to this word
            word[i] = backup;
        }
    }

    // Given wildcard to words, build the whole graph on words
    void BuildGraph(string& begin, string& end, vector<string>& wordList) {
        auto Process = [this](string& w) {
            auto wordId = GetId(w);
            for (size_t i = 0;i < w.size();i++) {
                char backup = w[i];
                w[i] = '*';
                for (int source : m_wildcardToWords[GetId(w)]) {
                    m_neighbors[source].push_back(wordId);
                }
                w[i] = backup;
            }
        };
        Process(begin);
        Process(end);
        for_each(wordList.begin(), wordList.end(), Process);
    }

    unordered_map<string, int> m_ids;
    unordered_map<int, vector<int>> m_wildcardToWords;
    unordered_map<int, vector<int>> m_neighbors;
};
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