<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>all the <strong>shortest transformation sequences</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words </em><code>[beginWord, s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>k</sub>]</code>.</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> [[&quot;hit&quot;,&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;cog&quot;],[&quot;hit&quot;,&quot;hot&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]]
<strong>Explanation:</strong>&nbsp;There are 2 shortest transformation sequences:
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; &quot;cog&quot;
&quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;lot&quot; -&gt; &quot;log&quot; -&gt; &quot;cog&quot;
</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> []
<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;= 5</code></li>
	<li><code>endWord.length == beginWord.length</code></li>
	<li><code>1 &lt;= wordList.length &lt;= 500</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>
	<li>The <strong>sum</strong> of all shortest transformation sequences does not exceed <code>10<sup>5</sup></code>.</li>
</ul>

#include <bits/stdc++.h>
using namespace std;

class Solution
{
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string> &wordList)
    {
        // Add beginWord to the word list
        wordList.push_back(beginWord);
        int n = wordList.size();
        
        // Find the index of the target word (endWord)
        unordered_map<string, int> wordToIndex;
        for (int i = 0; i < n; i++) {
            wordToIndex[wordList[i]] = i;
        }
        
        if (wordToIndex.find(endWord) == wordToIndex.end()) return {}; // If endWord not in wordList, return empty
        
        int target = wordToIndex[endWord];
        
        // Create adjacency list
        vector<vector<int>> adj(n);
        auto canTransform = [&](const string &a, const string &b) -> bool {
            int diff = 0;
            for (int i = 0; i < a.size(); i++) {
                if (a[i] != b[i]) diff++;
                if (diff > 1) return false;
            }
            return diff == 1;
        };
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (canTransform(wordList[i], wordList[j])) {
                    adj[i].push_back(j);
                    adj[j].push_back(i);
                }
            }
        }
        
        // BFS to find the shortest paths
        queue<vector<int>> q;
        vector<int> dist(n, INT_MAX);
        vector<vector<int>> parents(n); // Keep track of parents for backtracking
        
        int start = wordToIndex[beginWord];
        q.push({start});
        dist[start] = 0;
        
        while (!q.empty()) {
            auto path = q.front();
            q.pop();
            int node = path.back();
            
            if (node == target) break;
            
            for (int neighbor : adj[node]) {
                if (dist[neighbor] > dist[node] + 1) {
                    dist[neighbor] = dist[node] + 1;
                    parents[neighbor] = {node};
                    auto newPath = path;
                    newPath.push_back(neighbor);
                    q.push(newPath);
                } else if (dist[neighbor] == dist[node] + 1) {
                    parents[neighbor].push_back(node);
                }
            }
        }
        
        // Backtracking to find all shortest paths
        vector<vector<string>> result;
        vector<string> currentPath;
        
        function<void(int)> backtrack = [&](int node) {
            currentPath.push_back(wordList[node]);
            if (node == start) {
                vector<string> temp = currentPath;
                reverse(temp.begin(), temp.end());
                result.push_back(temp);
            } else {
                for (int parent : parents[node]) {
                    backtrack(parent);
                }
            }
            currentPath.pop_back();
        };
        
        if (dist[target] != INT_MAX) {
            backtrack(target);
        }
        
        return result;
    }
};

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