<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:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int n = wordList.size();
        unordered_map<string, vector<string>> graph;
        
        // Step 1: Build the graph with the correct adjacency logic
        for (int i = 0; i < n; i++) {
            string str1 = wordList[i];
            vector<string> temp;
            for (int j = 0; j < n; j++) {
                if (i == j) continue;
                string str2 = wordList[j];
                int diff = 0;
                
                // Compare character by character
                for (int k = 0; k < str1.size(); k++) {
                    if (str1[k] != str2[k]) diff++;
                    if (diff > 1) break; // Early exit if more than one character differs
                }
                
                if (diff == 1) temp.push_back(str2);  // Only add if exactly 1 character is different
            }
            graph[str1] = temp;
        }

        // Step 2: Check if the beginWord needs to be added to the graph
        if (graph.find(beginWord) == graph.end()) {
            vector<string> temp;
            for (int j = 0; j < n; j++) {
                string str2 = wordList[j];
                int diff = 0;
                
                for (int k = 0; k < beginWord.size(); k++) {
                    if (beginWord[k] != str2[k]) diff++;
                    if (diff > 1) break;
                }
                
                if (diff == 1) temp.push_back(str2);
            }
            graph[beginWord] = temp;
        }

        // Step 3: Return 0 if endWord is not in the wordList
        if (find(wordList.begin(), wordList.end(), endWord) == wordList.end()) {
            return 0;
        }

        // Step 4: BFS to find the shortest path
        unordered_map<string, int> dis;
        queue<string> q;
        
        for (int i = 0; i < n; i++) {
            dis[wordList[i]] = 1e9; // Initialize distances to infinity
        }
        dis[beginWord] = 1;
        q.push(beginWord);
        
        while (!q.empty()) {
            string str = q.front();
            q.pop();
            vector<string> neighbors = graph[str];
            
            for (string neighbor : neighbors) {
                if (dis[neighbor] > dis[str] + 1) {
                    dis[neighbor] = dis[str] + 1;
                    q.push(neighbor);
                }
            }
        }

        // Step 5: Return the distance to endWord, or 0 if not reachable
        return dis[endWord] == 1e9 ? 0 : dis[endWord];
    }

};
53175