<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:
bool differByOne(const std::string& word1, const std::string& word2) {
    int differences = 0;
    for (size_t i = 0; i < word1.length(); ++i) {
        if (word1[i] != word2[i]) {
            ++differences;
            if (differences > 1) {
                return false;
            }
        }
    }
    return differences == 1;
}
void createAdj(const vector<string>& wordList, vector<string> adj[]) {
    int n = wordList.size();

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (i != j && differByOne(wordList[i], wordList[j])) {
                adj[i].push_back(wordList[j]);
            }
        }
    }
}


    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int n = wordList.size();
        vector<string> adj[n];
        createAdj(wordList, adj);

    // Add the beginWord to the list and update the adjacency list
    std::vector<std::string> allWords = wordList;
    allWords.push_back(beginWord);
    std::vector<std::string> adjWithBegin[n + 1];
    createAdj(allWords, adjWithBegin);

    // BFS setup
    std::queue<std::pair<int, int>> q;
    std::vector<bool> visited(n + 1, false);

    // Find the index of the beginWord in the allWords list
    int beginIndex = n;  // last element is beginWord
    q.push({beginIndex, 1});
    visited[beginIndex] = true;

    // Perform BFS
    while (!q.empty()) {
        auto [currentIndex, level] = q.front();
        q.pop();

        // If we find the endWord, return the count
        if (allWords[currentIndex] == endWord) {
            return level;
        }

        // Iterate through the neighbors
        for (const auto& neighbor : adjWithBegin[currentIndex]) {
            int neighborIndex = std::distance(allWords.begin(), std::find(allWords.begin(), allWords.end(), neighbor));
            if (!visited[neighborIndex]) {
                visited[neighborIndex] = true;
                q.push({neighborIndex, level + 1});
            }
        }
    }

    // If endWord is not found, return 0
    return 0;
    }
};
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