<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>

struct Node{
    Node* child[26];
    bool eow;

    Node(){
        for(int i = 0; i < 26; i++){
            child[i] = NULL;
        }
        eow = false;
    }
};

class Solution {
public:
    Node* root;

    Solution() {
        root = new Node();
    }

    void insert(string s) {
        Node* node = root;
        for (int i = 0; i < s.size(); i++) {
            int idx = s[i] - 'a';
            if (node->child[idx] == NULL) {
                node->child[idx] = new Node();
            }
            node = node->child[idx];
        }
        node->eow = true;
    }

    bool solve(string s, int start, vector<int>& dp, int count) {
        Node* node = root;

        if (start == s.size()) return count >= 2; // Word formed by at least two smaller words

        if (dp[start] != -1) return dp[start];

        for (int i = start; i < s.size(); i++) {
            int idx = s[i] - 'a';

            if (node->child[idx] == NULL) {
                return dp[start] = false;
            }

            node = node->child[idx];

            if (node->eow) { // A valid subword is found
                if (solve(s, i + 1, dp, count + 1)) {
                    return dp[start] = true; // Return true if rest of the word can be split
                }
            }
        }

        return dp[start] = false; // No valid concatenation found
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        for (auto word : words) {
            insert(word); // Insert all words into the Trie
        }

        vector<string> ans;

        for (auto word : words) {
            vector<int> dp(word.size(), -1); // Memoization array for each word
            if (!word.empty() && solve(word, 0, dp, 0)) {
                ans.push_back(word); // If the word can be concatenated, add to result
            }
        }

        return ans;
    }
};

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