<p>Given an array of strings <code>words</code>, return <em>the smallest string that contains each string in</em> <code>words</code> <em>as a substring</em>. If there are multiple valid strings of the smallest length, return <strong>any of them</strong>.</p>

<p>You may assume that no string in <code>words</code> is a substring of another string in <code>words</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;alex&quot;,&quot;loves&quot;,&quot;leetcode&quot;]
<strong>Output:</strong> &quot;alexlovesleetcode&quot;
<strong>Explanation:</strong> All permutations of &quot;alex&quot;,&quot;loves&quot;,&quot;leetcode&quot; would also be accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;catg&quot;,&quot;ctaagt&quot;,&quot;gcta&quot;,&quot;ttca&quot;,&quot;atgcatc&quot;]
<strong>Output:</strong> &quot;gctaagttcatgcatc&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 12</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 20</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    string shortestSuperstring(vector<string>& A) {
        int n = A.size();
        vector<vector<string>> dp(1<<n, vector<string>(n));
        vector<vector<int>> overlap(n, vector<int>(n,0));

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if ( i!= j) {
                    for (int k = min(A[i].size(), A[j].size()); k > 0; --k)
                        if (A[i].substr(A[i].size()-k) == A[j].substr(0,k)) {
                            overlap[i][j] = k; 
                            break;
                        }
                }
            }
        }

        for(int i = 0; i < n; ++i) dp[1<<i][i] += A[i];

        for(int mask=1; mask<(1<<n); ++mask){
            for(int j = 0; j < n; ++j) 
                if ((mask&(1<<j)) > 0) {
                    for (int i = 0; i < n; ++i) 
                        if (i != j && (mask&(1<<i)) > 0) {
                            string tmp = dp[mask^(1<<j)][i] + A[j].substr(overlap[i][j]);

                            if(dp[mask][j].empty() || tmp.size() < dp[mask][j].size())
                                dp[mask][j] = tmp;
                        }
                }
        }

        int last = (1<<n)-1;
        string res = dp[last][0];

        for (int i = 1; i < n; ++i) 
            if (dp[last][i].size() < res.size())
                res = dp[last][i];

        return res;
    }
};
210400