<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:
                    # Calculate maximum overlap of words[i] and words[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 shortest superstring that contains all words in mask
        # and ends with words[i]
        dp = [[''] * n for _ in range(1 << n)]
        
        # Initialize dp with single words
        for i in range(n):
            dp[1 << i][i] = words[i]
        
        # Iterate over all subsets of words
        for mask in range(1 << n):
            for i in range(n):
                if not (mask & (1 << i)):
                    continue
                for j in range(n):
                    if mask & (1 << j):
                        continue
                    # Calculate new mask
                    new_mask = mask | (1 << j)
                    # Calculate new superstring
                    new_superstring = dp[mask][i] + words[j][overlap[i][j]:]
                    # Update dp table
                    if dp[new_mask][j] == '' or len(new_superstring) < len(dp[new_mask][j]):
                        dp[new_mask][j] = new_superstring
        
        # Find the shortest superstring from all possibilities
        min_superstring = None
        full_mask = (1 << n) - 1
        for i in range(n):
            if min_superstring is None or len(dp[full_mask][i]) < len(min_superstring):
                min_superstring = dp[full_mask][i]
        
        return min_superstring

# Example usage:
# solution = Solution()
# print(solution.shortestSuperstring(["alex","loves","leetcode"]))  # Output: "alexlovesleetcode"
# print(solution.shortestSuperstring(["catg","ctaagt","gcta","ttca","atgcatc"]))  # Output: "gctaagttcatgcatc"
20300