<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:
    unordered_map<string, bool> memo;

    bool concatWord(const string& w, const unordered_set<string>& dict) {
        auto it = memo.find(w);
        if (it != memo.end())
            return memo[w];

        for (int i = 1; i < w.size(); ++i) {
            string prefix = w.substr(0, i);
            string suffix = w.substr(i, w.size() - i);
            if (dict.count(prefix) == 0)
                continue;

            if (dict.count(suffix) > 0) {
                memo[w] = true;
                return true;
            }

            if (concatWord(suffix, dict)) {
                memo[w] = true;
                return true;
            }
        }

        memo[w] = false;
        return false;
    }

    // 24.08.2024
    vector<string> findAllConcatenatedWordsInADict1(vector<string>& words) {
        unordered_set<string> dict;

        for (auto w: words)
            dict.insert(w);

        vector<string> answer;

        for (auto w: words) {
            if (concatWord(w, dict))
                answer.push_back(w);
        }

        return answer;
    }

    struct TreeNode {
        unordered_map<char, TreeNode*> nodes;
        bool endOfWord = false;
    } root;


    void addWord(TreeNode* root, const string& word) {
        TreeNode* node = root;
        for (char c: word) {
            if (node->nodes.find(c) == node->nodes.end())
                node->nodes[c] = new TreeNode();
            
            node = node->nodes[c];
        }

        node->endOfWord = true;
    }

    void createTrie(vector<string>& words) {
        for (auto& w: words) {
            addWord(&root, w);
        }
    }

    unordered_map<string, int> memo2;

    int searchConCat(const string& word, int pos) {
        string wrd = word.substr(pos);
        if (memo2.find(wrd) != memo2.end())
            return memo2[wrd];

        if (pos == word.size())
            return 0;

        int found = 0;
        TreeNode* node = &root;
        for (int i = pos; i < word.size(); ++i) {
            if (node->nodes.find(word[i]) == node->nodes.end())
                break;

            char c = word[i];
            node = node->nodes[c];
            if (node->endOfWord) {
                int prefix_found = searchConCat(word, i + 1);
                if (prefix_found >= 0) {
                    found = 1 + prefix_found;
                    break;
                }
            }
        }

        memo2[wrd] = (found == 0) ? -1 : found;
        
        return memo2[wrd];
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        createTrie(words);

        vector<string> answer;

        for (auto w: words) {
            if (searchConCat(w, 0) > 1)
                answer.push_back(w);
        }

        return answer;
    }

};
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