<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>

#include <unordered_map>

class Solution {
public:
    
    map<string, bool> memo;

    bool solve(string& target, string& temp, vector<vector<string>>& adj) {
        if (temp.length() == target.length()) {
            return temp == target;
        }
        
        if (temp.length() > target.length()) return false;
        
        char start = target[temp.length()];

        if (memo.find(temp) != memo.end()) return memo[temp];

        for (auto& it : adj[start - 'a']) {
            string new_temp = temp + it;
            if (new_temp.length() > target.length()) continue;

            if(it==target.substr(temp.length(),it.length())){
                if (solve(target, new_temp, adj)) {
                memo[temp] = true;  
                return true;
            }
            }
        }

        memo[temp] = false;  
        return false;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        vector<vector<string>> adj(26);
        int n = words.size();

        for (int i = 0; i < n; i++) {
                adj[words[i][0] - 'a'].push_back(words[i]);
        }

        vector<string> ans;

        for (int i = 0; i < n; i++) {
            memo.clear(); 
            bool flag = false;

            for (auto& it : adj[words[i][0] - 'a']) {
                if (it != words[i] && it==words[i].substr(0,it.length())) {
                    string temp = it;
                    if (solve(words[i], temp, adj)) {
                        flag = true;
                        break;
                    }
                }
            }

            if (flag) ans.push_back(words[i]);
        }

        return ans;
    }
};

119676