<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 <queue>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

class Solution {
public:
    int re(unordered_map<string, vector<string>>& adjList, const string& start, const string& end) {
    if (start == end) return 0;
    
    unordered_map<string, int> distances;
    for (const auto& pair : adjList) {
        distances[pair.first] = numeric_limits<int>::max();
    }
    
    queue<string> q;
    q.push(start);
    distances[start] = 0;
    
    while (!q.empty()) {
        string current = q.front();
        q.pop();
        
        for (const string& neighbor : adjList[current]) {
            if (distances[neighbor] == numeric_limits<int>::max()) {
                distances[neighbor] = distances[current] + 1;
                q.push(neighbor);
                
                if (neighbor == end) {
                    return distances[neighbor];
                }
            }
        }
    }
    
    return -1;
}


   

    int ladderLength(string bw, string ew, vector<string>& wl) {
        unordered_map<string, vector<string>> h;
        unordered_map<string, bool> v;
        wl.push_back(bw);
        bool in = false;
        for (auto x : wl) {
            v[x] = false;
            if (x == ew) {
                in = true;
            }
            h[x] = {};
            for (auto y : wl) {
                int d = 0;
                bool f = true;
                for (int i = 0; i < y.size(); i++) {
                    if (y[i] != x[i]) {
                        d++;
                    }
                    if (d >= 2) {
                        f = false;
                        break;
                    }
                }
                if (f) {
                    h[y].push_back(x);
                    h[x].push_back(y);
                }
            }
        }
        if (!in) {
            return 0; // Return 0 as a ladder length of 0 indicates no
                      // transformation sequence
        }
        int k =re(h,bw,ew) ;
        return k!=0 ? k+1 : 0 ;
    }
};

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