<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 getDiff(string s1, string s2) {
        int ret = 0;
        for (int i = 0; i < s1.size(); i++) {
            ret += (s1[i] != s2[i]);
        }
        return ret;
    }
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_map<string, vector<string>> g;
        unordered_map<string, int> visited1, visited2;
        bool flag = false;
        visited1[beginWord] = 0;
        visited1[endWord] = 0;

        visited2[beginWord] = 0;
        visited2[endWord] = 0;
        for (int i = 0; i < wordList.size(); i++) {
            visited1[wordList[i]] = 0;
            visited2[wordList[i]] = 0;
            if (getDiff(wordList[i], beginWord) == 1) {
                g[beginWord].push_back(wordList[i]);
                g[wordList[i]].push_back(beginWord);
            }
            if (getDiff(wordList[i], endWord) == 1) {
                g[wordList[i]].push_back(endWord);
                g[endWord].push_back(wordList[i]);
            }
            if (endWord == wordList[i]) flag = 1;
        }
        if (flag == false) return 0;
        for (int i = 0; i < wordList.size(); i++) {
            if (wordList[i] == beginWord || wordList[i] == endWord) continue;
            for (int j = i + 1; j < wordList.size(); j++) {
                if (wordList[j] == beginWord || wordList[j] == endWord) continue;
                if (getDiff(wordList[i], wordList[j]) == 1) {
                    g[wordList[i]].push_back(wordList[j]);
                    g[wordList[j]].push_back(wordList[i]);
                }
            }
        }
        queue<string> q1, q2;
        q1.push(beginWord);
        q2.push(endWord);
        int ans = 0;
        while (!q1.empty() && !q2.empty()) {
            int sz1 = q1.size();
            int sz2 = q2.size();
            ans++;

            for (int i = 0; i < sz1; i++) {
                string cur = q1.front();
                string prev = q1.front();
                visited1[cur] = 1;
                q1.pop();
                // cout << "cur: " << cur << "\n";
                if (cur == endWord) return ans;
                for (auto &neighbor: g[cur]) {
                    // cout << "neighbor: " << neighbor << "\n";
                    if (visited1[neighbor] == 1) continue;
                    // cout << "neighbor2: " << neighbor << "\n";
                    q1.push(neighbor);
                }
            }
            for (int i = 0; i < sz2; i++) {
                string cur = q2.front();
                string prev = q2.front();
                visited2[cur] = 1;
                q2.pop();
                // cout << "cur: " << cur << "\n";
                if (cur == beginWord) return ans;
                for (auto &neighbor: g[cur]) {
                    // cout << "neighbor: " << neighbor << "\n";
                    if (visited2[neighbor] == 1) continue;
                    // cout << "neighbor2: " << neighbor << "\n";
                    q2.push(neighbor);
                }
            }
        }
        return 0;
    }
};
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