<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 <unordered_set>
#include <unordered_map>
#include <queue>
#include <string>
#include <list>
#include <algorithm>

using namespace std;

class Solution {
private:
    vector<vector<string>> allPaths; // List of paths from beginWord to endWord
    unordered_map<string, unordered_set<string>> predecessorsMap; // Map to track the predecessors of each word in the shortest paths

public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        allPaths.clear();
        unordered_set<string> wordSet(wordList.begin(), wordList.end()); // Convert word list to a set for efficient lookups
        if (wordSet.find(endWord) == wordSet.end()) {
            return allPaths; // If endWord is not in the word set, return empty list
        }
      
        wordSet.erase(beginWord); // Remove beginWord from the set to prevent cycles
        unordered_map<string, int> distanceMap; // Map to track the shortest path distances for words
        distanceMap[beginWord] = 0; // Distance from beginWord to itself is 0
        predecessorsMap.clear(); // Initialize the predecessors map
        queue<string> queue; // Queue for BFS
        queue.push(beginWord);
        bool isEndWordFound = false; // Flag to check if endWord is found
        int steps = 0; // Step counter for BFS

        while (!queue.empty() && !isEndWordFound) {
            ++steps;
            int levelSize = queue.size();
            for (int i = 0; i < levelSize; ++i) {
                string currentWord = queue.front();
                queue.pop();
                string originalWord = currentWord;

                for (int j = 0; j < currentWord.size(); ++j) {
                    char originalChar = currentWord[j];
                    for (char c = 'a'; c <= 'z'; ++c) { // Try all possible one-letter mutations
                        currentWord[j] = c;
                        if (distanceMap.find(currentWord) != distanceMap.end() && distanceMap[currentWord] == steps) {
                            predecessorsMap[currentWord].insert(originalWord);
                        }
                        if (wordSet.find(currentWord) == wordSet.end()) { 
                            continue; // If the new word isn't in the set, skip it
                        }
                        // Update distance map and predecessors map for new words
                        predecessorsMap[currentWord].insert(originalWord);
                        wordSet.erase(currentWord); // Remove new word to prevent revisiting
                        queue.push(currentWord);
                        distanceMap[currentWord] = steps;

                        if (endWord == currentWord) {
                            isEndWordFound = true; // Found the endWord; will finish after this level
                        }
                    }
                    currentWord[j] = originalChar; // Restore original character before next iteration
                }
            }
        }

        if (isEndWordFound) { // If the end word has been reached
            deque<string> path; // Path stack for reconstructing paths
            path.push_back(endWord);
            backtrackPath(path, beginWord, endWord); // Perform DFS to build all shortest paths
        }

        return allPaths; // Return the list of all shortest paths
    }

private:
    void backtrackPath(deque<string> path, const string& beginWord, const string& currentWord) {
        if (currentWord == beginWord) { // If the beginning of the path is reached, add it to allPaths
            allPaths.push_back(vector<string>(path.begin(), path.end()));
            return;
        }
      
        // Recursively go through all predecessors of the current word, adding them to the path
        for (const auto& predecessor : predecessorsMap[currentWord]) {
            path.push_front(predecessor); // Push the predecessor onto the path
            backtrackPath(path, beginWord, predecessor); // Continue backtracking
            path.pop_front(); // Remove the predecessor to backtrack
        }
    }
};

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