<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:
    const int p = 31;
    const int mod = 1e9 + 9;
    long long pwr[31];
    long long inv[31];
    long long mexp(int a, int b) {
        long long res = 1;
        while (b>0) {
            if(b&1) res = (res*a)%mod;
            a = (1LL*a*a)%mod;
            b = b>>1;
        }
        return res;
    }
    void precomp() {
        pwr[0] = 1, inv[0]=1;
        for (int i = 1; i < 31; ++i) {
            pwr[i] = (pwr[i - 1] * p) % mod;
        }
        for (int i = 1; i < 31; ++i) {
            inv[i] = mexp(pwr[i], mod - 2);
        }
    }
    void comp(string& s, vector<vector<long long>>& v,
              vector<unordered_set<long long>>& total_hash, int i) {
        int n = s.size();
        vector<long long> hash(n + 1);
        hash[0] = 0;
        for (int i = 1; i <= n; ++i) {
            hash[i] = ((hash[i - 1]) + (s[i - 1] - 'a' + 1) * pwr[i - 1]) % mod;
            // cout<<hash[i]<<endl;
        }
        v.push_back(hash);
        total_hash[n].insert(hash[n]);
    }
    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        int m = words.size();
        vector<vector<long long>> v;
        vector<unordered_set<long long>> total_hash(31);

        precomp();
        for (int i = 0; i < m; ++i) {
            comp(words[i], v, total_hash, i);
        }
        vector<string> ans;
        for (int k = 0; k < m; ++k) {
            int n = words[k].size();
            vector<bool> dp(n + 1, 0);
            dp[0] = 1;
            queue<int> q;
            q.push(0);
            while (!q.empty()) {
                // if(k==2)    cout<<1<<endl;
                if(dp[n]==1)    break;
                int ind = q.front();
                q.pop();
                for (int j = ind + 1; j <= n&&(j-ind<n); ++j) {
                    if(dp[j]==1)    continue;
                    int sz = j - ind;
                    int curr = (v[k][j] - v[k][ind] + mod) % mod;
                    if(k==2){
                        // cout<<ind<<" "<<j<<" "<<curr<<endl;
                    }
                    curr = (curr*inv[ind])%mod;
                    if(total_hash[sz].count(curr)){
                        dp[j]=1;
                        q.push(j);
                    }
                }
            }
            if(dp[n]==1){
                ans.push_back(words[k]);
            }
        }
        return ans;
    }
};
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