Find the Shortest Superstring
<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>
Bit Manipulation; Array; String; Dynamic Programming; Bitmask
class Solution {
    public String shortestSuperstring(String[] words) {
        int n = words.length;
        int[][] g = new int[n][n];
        for (int i = 0; i < n; ++i) {
            String a = words[i];
            for (int j = 0; j < n; ++j) {
                String b = words[j];
                if (i != j) {
                    for (int k = Math.min(a.length(), b.length()); k > 0; --k) {
                        if (a.substring(a.length() - k).equals(b.substring(0, k))) {
                            g[i][j] = k;
                            break;
                        }
                    }
                }
            }
        }
        int[][] dp = new int[1 << n][n];
        int[][] p = new int[1 << n][n];
        for (int i = 0; i < 1 << n; ++i) {
            Arrays.fill(p[i], -1);
            for (int j = 0; j < n; ++j) {
                if (((i >> j) & 1) == 1) {
                    int pi = i ^ (1 << j);
                    for (int k = 0; k < n; ++k) {
                        if (((pi >> k) & 1) == 1) {
                            int v = dp[pi][k] + g[k][j];
                            if (v > dp[i][j]) {
                                dp[i][j] = v;
                                p[i][j] = k;
                            }
                        }
                    }
                }
            }
        }
        int j = 0;
        for (int i = 0; i < n; ++i) {
            if (dp[(1 << n) - 1][i] > dp[(1 << n) - 1][j]) {
                j = i;
            }
        }
        List<Integer> arr = new ArrayList<>();
        arr.add(j);
        for (int i = (1 << n) - 1; p[i][j] != -1;) {
            int k = i;
            i ^= (1 << j);
            j = p[k][j];
            arr.add(j);
        }
        Set<Integer> vis = new HashSet<>(arr);
        for (int i = 0; i < n; ++i) {
            if (!vis.contains(i)) {
                arr.add(i);
            }
        }
        Collections.reverse(arr);
        StringBuilder ans = new StringBuilder(words[arr.get(0)]);
        for (int i = 1; i < n; ++i) {
            int k = g[arr.get(i - 1)][arr.get(i)];
            ans.append(words[arr.get(i)].substring(k));
        }
        return ans.toString();
    }
}