<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 <vector>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <climits>

using namespace std;

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        wordList.push_back(beginWord);
        int n = wordList.size();
        
        // Create word-to-index map for easy reference
        unordered_map<string, int> wordIndex;
        for (int i = 0; i < n; i++) {
            wordIndex[wordList[i]] = i;
        }

        // Floyd-Warshall to calculate shortest path between all pairs of words
        vector<vector<int>> dist(n, vector<int>(n, INT_MAX));
        for (int i = 0; i < n; i++) {
            dist[i][i] = 0;
        }
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isOneLetterDiff(wordList[i], wordList[j])) {
                    dist[i][j] = dist[j][i] = 1;
                }
            }
        }

        // Running Floyd-Warshall algorithm to compute shortest paths
        for (int k = 0; k < n; k++) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    if (dist[i][k] < INT_MAX && dist[k][j] < INT_MAX) {
                        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        }

        // If there's no path from beginWord to endWord
        if (wordIndex.find(endWord) == wordIndex.end() || dist[wordIndex[beginWord]][wordIndex[endWord]] == INT_MAX) {
            return {};
        }

        // DFS to find all shortest paths using the heuristic
        vector<vector<string>> results;
        vector<string> path = {beginWord};
        dfs(wordIndex[beginWord], wordIndex[endWord], dist, wordList, path, results);
        return results;
    }

private:
    bool isOneLetterDiff(const string& word1, const string& word2) {
        int diff = 0;
        for (int i = 0; i < word1.size(); i++) {
            if (word1[i] != word2[i]) {
                diff++;
                if (diff > 1) return false;
            }
        }
        return diff == 1;
    }

    void dfs(int current, int target, vector<vector<int>>& dist, vector<string>& wordList, vector<string>& path, vector<vector<string>>& results) {
        if (current == target) {
            results.push_back(path);
            return;
        }

        for (int next = 0; next < wordList.size(); next++) {
            if (dist[current][next] == 1 && dist[next][target] == dist[current][target] - 1) {
                path.push_back(wordList[next]);
                dfs(next, target, dist, wordList, path, results);
                path.pop_back();
            }
        }
    }
};

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