<p>Given an array of strings <code>words</code> (<strong>without duplicates</strong>), return <em>all the <strong>concatenated words</strong> in the given list of</em> <code>words</code>.</p>

<p>A <strong>concatenated word</strong> is defined as a string that is comprised entirely of at least two shorter words (not necessarily distinct)&nbsp;in the given array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;cats&quot;,&quot;catsdogcats&quot;,&quot;dog&quot;,&quot;dogcatsdog&quot;,&quot;hippopotamuses&quot;,&quot;rat&quot;,&quot;ratcatdogcat&quot;]
<strong>Output:</strong> [&quot;catsdogcats&quot;,&quot;dogcatsdog&quot;,&quot;ratcatdogcat&quot;]
<strong>Explanation:</strong> &quot;catsdogcats&quot; can be concatenated by &quot;cats&quot;, &quot;dog&quot; and &quot;cats&quot;; 
&quot;dogcatsdog&quot; can be concatenated by &quot;dog&quot;, &quot;cats&quot; and &quot;dog&quot;; 
&quot;ratcatdogcat&quot; can be concatenated by &quot;rat&quot;, &quot;cat&quot;, &quot;dog&quot; and &quot;cat&quot;.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;dog&quot;,&quot;catdog&quot;]
<strong>Output:</strong> [&quot;catdog&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> consists of only lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
	<li><code>1 &lt;= sum(words[i].length) &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        struct TrieNode {
            char c = '\0';
            bool isEnd = false;
        };
        vector<TrieNode> nodes;
        vector<vector<size_t>> edges;

        nodes.push_back(TrieNode{});
        edges.emplace_back();
        const size_t rootNode = 0;

        // build a trie
        for (const string& word : words) {
            size_t curr = rootNode;
            for (const char c : word) {
                // try to transition
                size_t next = curr;
                for (const size_t dst : edges[curr]) {
                    if (nodes[dst].c == c) {
                        next = dst;
                        break;
                    }
                }

                // if the transition failed, create the node
                if (next == curr) {
                    next = nodes.size();
                    nodes.push_back(TrieNode{.c = c});
                    edges[curr].push_back(next);
                    edges.emplace_back();
                }

                curr = next;
            }

            // mark the end of the word
            nodes[curr].isEnd = true;
        }

        unordered_set<string_view> result;
        unordered_map<string_view, bool> cache;
        const auto dfs = [&](const string_view word, size_t prev, size_t depth,
                             const auto& recFun) {
            if (word.empty()) {
                return true;
            }

            for (size_t pos = 0; pos < word.length(); ++pos) {
                // try to find the corresponding node
               // cout << "At " << nodes[prev].c << endl;
                //cout << "Looking for " << word[pos] << endl;

                size_t curr = prev;
                for (const size_t dst : edges[curr]) {
                    if (nodes[dst].c == word[pos]) {
                        curr = dst;
                        break;
                    }
                }

                if (curr == prev) {
                    //cout << "No way" << endl;
                    return false;
                }

                // Try to split at the current position
                if (nodes[curr].isEnd) {
                    //cout << "Trying to split " << endl;
                    const auto suffix = string_view(word).substr(pos + 1);
                    bool isConcat = false;
                    if (const auto hit = cache.find(suffix); hit != cache.end()) {
                        isConcat = hit->second;
                    } else {
                        isConcat = cache[suffix] = recFun(suffix, rootNode, depth + 1, recFun);
                    }
                    if (isConcat) {
                        //cout << "Split success: " << suffix << endl;
                        if (depth == 0 && !suffix.empty()) {
                            //cout << "Add " << word << endl;
                            result.insert(word);
                        }
                        return true;
                    }
                }

                //cout << "Split fail " << endl;
                prev = curr;
            }

            return false;
        };

        for (const string& word : words) {
            dfs(word, rootNode, 0, dfs);
        }

        return vector<string>(result.begin(), result.end());
    }
};
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