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:
    def shortestSuperstring(self, words: List[str]) -> str:
        n = len(words)
        g = [[0] * n for _ in range(n)]
        for i, a in enumerate(words):
            for j, b in enumerate(words):
                if i != j:
                    for k in range(min(len(a), len(b)), 0, -1):
                        if a[-k:] == b[:k]:
                            g[i][j] = k
                            break
        dp = [[0] * n for _ in range(1 << n)]
        p = [[-1] * n for _ in range(1 << n)]
        for i in range(1 << n):
            for j in range(n):
                if (i >> j) & 1:
                    pi = i ^ (1 << j)
                    for k in range(n):
                        if (pi >> k) & 1:
                            v = dp[pi][k] + g[k][j]
                            if v > dp[i][j]:
                                dp[i][j] = v
                                p[i][j] = k
        j = 0
        for i in range(n):
            if dp[-1][i] > dp[-1][j]:
                j = i
        arr = [j]
        i = (1 << n) - 1
        while p[i][j] != -1:
            i, j = i ^ (1 << j), p[i][j]
            arr.append(j)
        arr = arr[::-1]
        vis = set(arr)
        arr.extend([j for j in range(n) if j not in vis])
        ans = [words[arr[0]]] + [words[j][g[i][j] :] for i, j in pairwise(arr)]
        return ''.join(ans)
