<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:
    string delimit = "";
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_map<string, vector<string>> adj;
        
        // Build adjacency list

        int endExists = -1;
        for(int i=wordList.size() - 1;i>=0;i--) {
            if(wordList[i] == endWord) {
                endExists = i;
            }

            int diff = wordDiff(wordList[i], beginWord);
            if(diff == 1) {
                adj[wordList[i]].push_back(beginWord);
            }

            for(int j=i-1;j>=0;j--) {
                diff = wordDiff(wordList[i], wordList[j]);
                if(diff == 1) {
                    adj[wordList[i]].push_back(wordList[j]);
                    adj[wordList[j]].push_back(wordList[i]);
                }
            }
        } 

        int answer = 1;

        // If end does not exist
        if(endExists == -1) {
            return 0;
        }

        // BFS 
        queue<string> next;
        set<string> visited;

        next.push(endWord);
        next.push(delimit);

        while(!next.empty()) {
            string current = next.front();
            next.pop();
            if(current == delimit) {
                if(!next.empty()) {
                    next.push(delimit);
                    answer++;
                }
            } else {
                visited.insert(current);
                if(current == beginWord) {
                    return answer;
                }
                for(int i=0;i<adj[current].size();i++) {
                    if(!visited.count(adj[current][i])) {
                        next.push(adj[current][i]);
                    }
                }
            }
        }

        return 0;
    }

    int wordDiff(string& s1, string& s2) {
        int ptr = 0;
        int answer = 0;
        while(ptr < s1.size()) {
            answer += s1[ptr] != s2[ptr];
            ptr++;
        }
        return answer;
    }
};
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