<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>

typedef vector <string> vs;
typedef vector <vs> vvs;
class Solution {
public:
    void lesser(string &ret, string s) {
        if (ret.size() > s.size())
            ret = s;
    }
    string shortestSuperstring(vector<string>& a) {
        int n = a.size();
        vvs dp(1 << n, vs(n, string(241, 'a'))); 
        vvs dist(n, vs(n, ""));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) continue ;
                dist[i][j] = a[j];
                for (int ii = 0; ii < a[i].size(); ii++) {
                    if (ii + a[j].size() < a[i].size()) continue ;
                    int found = true;
                    for (int jj = 0, kk = ii; kk < a[i].size(); jj++, kk++)
                        if (a[i][kk] != a[j][jj]) {
                            found = false;
                            break ;
                        }
                    if (found == true) {
                        dist[i][j] = a[j].substr(a[i].size() - ii);
                        break ;
                    }
                }
            }
        }
        dp[0][0] = "";
        for (int i = 0; i < n; i++)
            dp[1 << i][i] = a[i];
        for (int i = 1; i < (1 << n); i++) {
            if (__builtin_popcount(i) == 1) {
                continue ;
            }
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) > 0) {
                    int prev_mask = i ^ (1 << j);
                    // we are going to add j to make this work
                    for (int k = 0; k < n; k++) {
                        if (j == k) continue ;
                        if (dp[i][j].size() > dp[prev_mask][k].size() + dist[k][j].size())
                            dp[i][j] = dp[prev_mask][k] + dist[k][j];   
                    }
                }
            }
        }
        string ret = string(300, 'a');
        for (int i = 0; i < n; i++) {
            lesser(ret, dp[(1 << n) - 1][i]);
        }
        return ret;
    }
};
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