<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>

from typing import List

class Solution:
    def shortestSuperstring(self, words: List[str]) -> str:
        n = len(words)
        
        # Precompute the overlap between each pair of words
        overlap = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                if i != j:
                    m = min(len(words[i]), len(words[j]))
                    for k in range(m, 0, -1):
                        if words[i].endswith(words[j][:k]):
                            overlap[i][j] = k
                            break
        
        # dp[mask][i] will store the minimum length of superstring that contains
        # all words in the subset represented by mask, ending with words[i]
        dp = [[float('inf')] * n for _ in range(1 << n)]
        parent = [[-1] * n for _ in range(1 << n)]
        
        # Initialize dp for single words
        for i in range(n):
            dp[1 << i][i] = len(words[i])
        
        # Fill the dp table
        for mask in range(1 << n):
            for i in range(n):
                if mask & (1 << i):
                    for j in range(n):
                        if mask & (1 << j) == 0:
                            new_mask = mask | (1 << j)
                            new_len = dp[mask][i] + len(words[j]) - overlap[i][j]
                            if new_len < dp[new_mask][j]:
                                dp[new_mask][j] = new_len
                                parent[new_mask][j] = i
        
        # Find the minimum length superstring
        min_len = float('inf')
        last = -1
        final_mask = (1 << n) - 1
        for i in range(n):
            if dp[final_mask][i] < min_len:
                min_len = dp[final_mask][i]
                last = i
        
        # Reconstruct the shortest superstring
        mask = final_mask
        perm = []
        while last != -1:
            perm.append(last)
            next_last = parent[mask][last]
            mask ^= (1 << last)
            last = next_last
        
        perm.reverse()
        
        # Build the result string
        superstring = words[perm[0]]
        for i in range(1, len(perm)):
            overlap_len = overlap[perm[i-1]][perm[i]]
            superstring += words[perm[i]][overlap_len:]
        
        return superstring
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