<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 {
    unordered_map<string, int> ids;
    unordered_map<int, string> id_str;
    using graph_t = unordered_map<int, vector<int>>;

    // return id of the word, if it doesn't exists, it will create one
    int getId(const string &word) {
        auto res = ids.emplace(word, ids.size());
        if (res.second) {
            // new id
            id_str[res.first->second] = word;
        }
        return res.first->second;
    }

    // transform word into special words, and put them in the graph
    void transform(string word, graph_t &graph) {
        int word_id = getId(word);
        int n = word.size();

        for (int i = 0; i < n; i++) {
            char bak = word[i];
            word[i] = '*';
            int transformed_id = getId(word);
            word[i] = bak;

            // put the link into the graph
            graph[word_id].push_back(transformed_id);
            graph[transformed_id].push_back(word_id);
        }
    }

    void reverseDfs(vector<vector<int>> &tree, vector<string> &buffer, int id,
                    vector<vector<string>> &output) {
        buffer.push_back(id_str[id]);
        if (tree[id].empty()) {
            // leaf
            output.emplace_back();
            for (auto riter = buffer.rbegin();; riter += 2) {
                output.back().push_back(*riter);
                if (riter == prev(buffer.rend()))
                    break;
            }
            buffer.pop_back();
            return;
        }
        for (auto child : tree[id]) {
            reverseDfs(tree, buffer, child, output);
        }
        buffer.pop_back();
    }

  public:
    vector<vector<string>> findLadders(string beginWord, string endWord,
                                       vector<string> &wordList) {
        graph_t graph;
        // preparation
        for (auto &word : wordList)
            transform(word, graph);
        if (!ids.count(beginWord)) {
            transform(beginWord, graph);
        }

        int start_id = getId(beginWord);
        auto iter = ids.find(endWord);
        int end_id;
        if (iter == ids.end())
            return {}; // end doesn't exists
        else
            end_id = iter->second;

        // Modified Dijkstra
        vector<int> dist(ids.size()); // id -> dist
        vector<vector<int>> prev(ids.size());
        set<pair<int, int>> pq; // dist -> id

        // this will update pq so that it contains correct dist
        // REQUIRES dist to contain old distance
        auto PQdecreaseKey = [&](int id, int new_dist) {
            pq.erase(make_pair(dist[id], id));
            pq.emplace(make_pair(new_dist, id));
        };
        auto PQextractPop = [&]() {
            int u = pq.begin()->second;
            pq.erase(pq.begin());
            return u;
        };
        auto PQcheck = [&](int id) {
            return pq.count(make_pair(dist[id], id));
        };

        for (int i = 0; i < ids.size(); i++) {
            dist[i] = INT_MAX;
            pq.emplace(INT_MAX, i);
        }
        PQdecreaseKey(start_id, 0);
        dist[start_id] = 0;

        while (!pq.empty()) {
            int u = PQextractPop();
            if (u == end_id || dist[u] == INT_MAX)
                break; // done
            for (auto v : graph[u]) {
                if (!PQcheck(v)) {
                    continue;
                }
                int alt = dist[u] + 1;
                if (alt < dist[v]) {
                    PQdecreaseKey(v, alt);
                    dist[v] = alt;
                    prev[v].push_back(u);
                } else if (alt == dist[v]) {
                    prev[v].push_back(u);
                }
            }
        }
        if (prev[end_id].empty())
            return {};

        vector<vector<string>> res;
        vector<string> buffer;
        reverseDfs(prev, buffer, end_id, res);

        // reverse DFS construct all paths
        return res;
    }
};
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