<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:
    /*
        Non-directional graph.
        BFS
        // Method 1
        unordered_set<string> remaining nodes
        time: N*(N-1)*...*3*2*1 = N!
        space: O(N)

        // Method 2
        build a graph:
        std::ordered_map<string, std::unordered_set<string>> neighbors;
        time: O(N+ N-1 + N-2 + ... + 3 + 2 +1) = O(N^2)
        space: O(N^2)
    */
    int ladderLength(string beginWord, string endWord,
                     vector<string>& wordList) {
        unordered_map<string, unordered_set<string>> graph =
            BuildGraph(wordList);
        for (const auto& word : wordList) {
            if (IsNeighbor(beginWord, word)) {
                graph[beginWord].insert(word);
            }
        }

        queue<string> queue_bfs;
        queue_bfs.push(beginWord);
        int length = 0;
        unordered_set<string> visited;
        while (!queue_bfs.empty()) {
            ++length;
            int size_batch = queue_bfs.size();
            for (int i = 0; i < size_batch; ++i) {
                if (queue_bfs.front() == endWord) {
                    return length;
                }
                const std::string front = queue_bfs.front();
                queue_bfs.pop();
                visited.insert(front);
                for (const auto neigh : graph[front]) {
                    if (visited.count(neigh) ==0) {
                        queue_bfs.push(neigh);
                    }
                }
            }
        }
        return 0;
    }

private:
    bool IsNeighbor(std::string_view word_1, std::string_view word_2) {
        int diff_count = 0;
        for (int i = 0; i < word_1.length(); ++i) {
            if (word_1[i] != word_2[i]) {
                ++diff_count;
            }
        }
        return diff_count == 1;
    }

    unordered_map<string, unordered_set<string>>
    BuildGraph(const vector<string>& wordList) {
        unordered_map<string, unordered_set<string>> result;
        for (int i = 0; i < wordList.size(); ++i) {
            for (int j = i + 1; j < wordList.size(); ++j) {
                if (IsNeighbor(wordList[i], wordList[j])) {
                    result[wordList[i]].insert(wordList[j]);
                    result[wordList[j]].insert(wordList[i]);
                }
            }
        }
        return result;
    }
};
55775