<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>

#define pii pair<int,int>
class Solution {
public:
    int ladderLength(string a, string b, vector<string>& v) {
         ios_base::sync_with_stdio(0), cin.tie(0);

         // If the end word is not in the word list, return 0 (no possible transformation)
         if (find(v.begin(), v.end(), b) == v.end()) {
             return 0;
         }

         // Add the start word to the word list
         v.push_back(a);
         int n = v.size();

         // Lambda to check if two words differ by exactly one character
         function<bool(string, string)> f = [&](string x, string y) -> bool {
             int cnt = 0;
             for (int i = 0; i < x.length(); i++) {
                 if (x[i] != y[i])
                     cnt++;
                 if (cnt > 1)
                     return false;
             }
             return cnt == 1;
         };

         // Find the target index for the end word
         int target = -1;
         vector<vector<int>> adj(n, vector<int>());

         // Build adjacency list for the word graph
         for (int i = 0; i < n; i++) {
             for (int j = i + 1; j < n; j++) {
                 if (f(v[i], v[j])) {
                     adj[i].push_back(j);
                     adj[j].push_back(i);
                 }
             }
             if (v[i] == b)
                 target = i;
         }

         // If no valid transformation can be made
         if (target == -1)
             return 0;

         // Perform BFS to find the shortest transformation sequence
         vector<int> dist(n, 1e9);
         queue<int> q;
         q.push(n - 1);  // Start from the last word (which is the `beginWord`)
         dist[n - 1] = 1;

         while (!q.empty()) {
             int node = q.front();
             q.pop();

             for (int neighbor : adj[node]) {
                 if (dist[neighbor] == 1e9) { // If this node hasn't been visited
                     dist[neighbor] = dist[node] + 1;
                     if (neighbor == target) {
                         return dist[neighbor];
                     }
                     q.push(neighbor);
                 }
             }
         }

         // If no path was found, return 0
         return 0;
    }
};

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