<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 {
    // O(26W) time
    vector<string> findWordsDifferChar(string word, unordered_set<string>& list) {
        vector<string> ans;
        
        for (int i = 0; i < word.length(); i++) {
            char orig = word[i];
            for (char c = 'a'; c <= 'z'; c++) {
                if (c != orig) {                
                    word[i] = c;
                    if (list.count(word)) {
                        ans.push_back(word);
                    }
                }
            }

            // restore
            word[i] = orig;
        }
        
        return ans;
    }
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        queue<string> q;
        unordered_set<string> seen;
        unordered_set<string> dict(wordList.begin(), wordList.end());

        if (dict.find(endWord) == dict.end()) {
            return 0;
        }
        
        // dict.insert (endWord);
        q.push(beginWord);
        seen.insert(beginWord);
        /* in case end is not in list at first */
        int dist = 1;
        while (!q.empty()) {
            int cnt = q.size();
            
            for (int i = 0; i < cnt; i++) {
                string curr = q.front(); q.pop();
                vector<string> adjWords = findWordsDifferChar(curr, dict);
                for (auto x : adjWords) {
                    /* we found the target */
                    if (x == endWord) {
                        return dist + 1;
                    }
                    if (!seen.count(x)) {
                        q.push(x);
                        seen.insert(x);
                    }
                }
            }
            
            /* go to next level */
            dist++;
        }
        
        return 0;
    }
};
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