<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:
    #define vi vector<int>
    #define vvi vector<vi>
    #define pb push_back
    #define dbg(x) cout << #x << " : " << x << endl

    int node_cnt = 1;
    int cnt[300005];
    int ends[300005];
    void add(string &s , vvi &trie){
        int x = 0;
        for(int i=0;i<s.size();i++){
            if(trie[x][s[i] - 'a'] == 0){
                trie.pb(vi(26 , 0));
                trie[x][s[i] - 'a'] = node_cnt;
                node_cnt++;
                assert(node_cnt == trie.size());
            }    
            x = trie[x][s[i] - 'a'];
            cnt[x]++;
            if(i == s.size() - 1){
                ends[x] = 1;
            }
        }
    }

    void erase(string &s , vvi &trie){
        int x = 0;
        for(int i=0;i<s.size();i++){
            assert(trie[x][s[i] - 'a'] != 0);

            x = trie[x][s[i] - 'a'];
            cnt[x]--;
            if(i == s.size() - 1){
                ends[x] = 0;
            }
        }
    }
    
    int dp[300005];
    int query(int pos , string &s , vvi &trie){
        if(pos == s.size())return 1;
        if(dp[pos] != -1)return dp[pos];

        int x = 0;
        int ok = 0;
        for(int i=pos;i<s.size();i++){
            if(cnt[trie[x][s[i] - 'a']] > 0 && trie[x][s[i] - 'a'] != 0){
                x = trie[x][s[i] - 'a'];
                if(ends[x] == 1){
                    // [pos...i] is a word
                    dbg(s.substr(pos , i-pos+1));
                    if(query(i+1 , s , trie) == 1)ok = 1;
                }
            }    
            else{
                break;
            }
        }
        return dp[pos] = ok;
    }

    vector<string> findAllConcatenatedWordsInADict(vector<string>& a) {
        // easy trie question
        // dynamic trie in strings, else TLE/MLE!

        cout << "a : ";
        for(string s : a){
            cout << s << " ";
        }
        cout << endl;

        memset(cnt , 0 , sizeof(cnt));
        memset(ends , 0 , sizeof(ends));
        memset(dp , -1 , sizeof(dp));

        vvi trie;
        node_cnt = 1;
        trie.pb(vi(26 , 0));

        vector<string> ans;
        int n = a.size();
        for(int i=0;i<n;i++){
            add(a[i] , trie);
        }
        for(int i=0;i<n;i++){
            erase(a[i] , trie);
            for(int j=0;j<=a[i].size();j++)dp[j] = -1;
            if(query(0 , a[i] , trie) == 1)ans.pb(a[i]);
            add(a[i] , trie);
        }
        return ans;
    }
};
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