<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 functools import lru_cache

class Solution:
    def shortestSuperstring(self, words: List[str]) -> str:
        n = len(words)
        
        # Calculate the overlap between each pair of words
        def calc_overlap(a, b):
            max_overlap = 0
            for i in range(1, min(len(a), len(b)) + 1):
                if a[-i:] == b[:i]:
                    max_overlap = i
            return max_overlap
        
        overlaps = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                if i != j:
                    overlaps[i][j] = calc_overlap(words[i], words[j])
        
        # Use dynamic programming to find the shortest superstring
        @lru_cache(None)
        def dp(mask, last):
            if mask == (1 << n) - 1:
                return ""
            res = None
            for i in range(n):
                if mask & (1 << i) == 0:
                    candidate = words[i][overlaps[last][i]:] + dp(mask | (1 << i), i)
                    if res is None or len(candidate) < len(res):
                        res = candidate
            return res
        
        # Try each word as the starting point
        result = None
        for i in range(n):
            candidate = words[i] + dp(1 << i, i)
            if result is None or len(candidate) < len(result):
                result = candidate
        
        return result
77000