<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>

class Solution {
public:
   vector<int>* precursor;
	vector<vector<string>> res;
	bool isANeighbor(string& s1, string& s2) {
		bool hasChanged = false;
		for (int i = 0; i < s1.size(); i++) {
			if (s1[i] != s2[i]) {
				if (hasChanged)
					return false;
				else
					hasChanged = true;
			}
		}
		return true;
	}
	void generateRoute(vector<string> right, vector<int>& precursor2, vector<string>& wordList) {
		if (precursor2.size() == 0) {
			res.push_back(right);
			return;
		}
		vector<string> copy;
		for (int i = 0; i < precursor2.size(); i++) {
			copy = right;
			// insert before the begin() cause we are back-tracing.
			copy.insert(copy.begin(), wordList[precursor2[i]]);
			generateRoute(copy, precursor[precursor2[i]], wordList);
		}
	
	}
	vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
		wordList.push_back(beginWord);
		int size = wordList.size();
		vector<int>* neighbors = new vector<int>[size];
		int ewordindex = -1;
		for (int i = 0; i < size; i++) {
			if (wordList[i] == endWord)
				ewordindex = i;
			for (int j = i + 1; j < size; j++) {
				if (isANeighbor(wordList[i],wordList[j])) {
					neighbors[i].push_back(j);
					neighbors[j].push_back(i);
				}
			}
		}
		vector<int> steps(size);//memorize distances
		queue<int> line;//BFS
		steps[size - 1] = 1;
		line.push(size - 1);//add the index of the beginWord to the queue
		precursor = new vector<int>[size];
		while (!line.empty()) {
			int pos = line.front();
			line.pop();
			if (wordList[pos] == endWord)
				//we stop here because we are looking for the shortest routes
				break;
			for (int i = 0; i < neighbors[pos].size(); i++) {
				if (steps[neighbors[pos][i]]==0) {
					//we have never visited here
					steps[neighbors[pos][i]] = steps[pos] + 1;
					precursor[neighbors[pos][i]].push_back(pos);
					line.push(neighbors[pos][i]);
				}
				else if (steps[neighbors[pos][i]] == steps[pos] + 1)
					//there's another route to get here and it has the SAME distance
					precursor[neighbors[pos][i]].push_back(pos);
			}
		
		}
		if (ewordindex == -1 || steps[ewordindex] == 0)
			return res;
		vector<string> right{endWord};
		generateRoute(right,precursor[ewordindex] , wordList);
		return res;
    }
};
14100