<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>

#include <algorithm>
#include <unordered_set>
#include <deque>

class Solution {
private:
    bool adjacent(string s1, string s2) {
        int diff = 0;
        for (int i=0; i<s1.size(); i++) {
            if (s1[i] != s2[i]) {
                diff++;
                if (diff > 1) return false;
            }
        }
        return diff <= 1;
    }
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int beginIndex = -1, endIndex = -1;
        for (int i=0; i<wordList.size(); i++) {
            beginIndex = (wordList[i] == beginWord) ? i : beginIndex;
            endIndex = (wordList[i] == endWord) ? i : endIndex;
        }
        if (endIndex == -1) return 0;  // endWord not in wordList;
        if (endWord.length() == 1) return 2;

        // build graph
        vector<unordered_set<int>> adjacency(wordList.size());
        for (int i=0; i<wordList.size(); i++) {
            for (int j=i+1; j<wordList.size(); j++) {
                if (adjacent(wordList[i], wordList[j])) {
                    adjacency[i].insert(j);
                    adjacency[j].insert(i);
                }
            }
        }

        // get initial words
        int transformations = 0;
        unordered_set<int> entryPoints;
        if (beginIndex == -1) {
            for (int i=0; i<wordList.size(); i++) {
                if (adjacent(beginWord, wordList[i])) entryPoints.insert(i);
            }
            transformations++;
        } else {
            entryPoints = {beginIndex};
        }

        // bfs
        unordered_set<int> visited;
        deque<int> queue;
        queue.insert(queue.end(), entryPoints.begin(), entryPoints.end());
        while (!queue.empty()) {
            int size = queue.size();
            transformations++;
            for (int i=0; i<size; i++) {
                int curr = queue.front();
                queue.pop_front();
                if (curr == endIndex) return transformations;
                if (visited.find(curr) != visited.end()) continue;
                visited.insert(curr);
                for (int adjacent : adjacency[curr]) {
                    if (visited.find(adjacent) != visited.end()) continue;
                    queue.push_back(adjacent);
                }
            }
        }
        return 0;
    }
};
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