<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)
        # Calculate 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:
                    # Check each possible overlap length from max to min
                    for l in range(min(len(words[i]), len(words[j])), 0, -1):
                        if words[i][-l:] == words[j][:l]:
                            overlap[i][j] = l
                            break
        
        # DP table, dp[mask][i] means the minimum length of superstring
        # formed by words represented by `mask` and ending with word i
        dp = [[float('inf')] * n for _ in range(1 << n)]
        parent = [[-1] * n for _ in range(1 << n)]
        
        # Initialize for single word cases
        for i in range(n):
            dp[1 << i][i] = len(words[i])
        
        # Build the dp table
        for mask in range(1 << n):
            for i in range(n):
                if mask & (1 << i):
                    # Previous mask, before adding word[i]
                    prev_mask = mask ^ (1 << i)
                    if prev_mask == 0:
                        continue
                    for j in range(n):
                        if prev_mask & (1 << j):
                            # Forming current from prev by adding word i after word j
                            cand = dp[prev_mask][j] + len(words[i]) - overlap[j][i]
                            if cand < dp[mask][i]:
                                dp[mask][i] = cand
                                parent[mask][i] = j
        
        # Find the minimum length superstring from all possible ending words
        min_len = float('inf')
        last = -1
        for i in range(n):
            if dp[(1 << n) - 1][i] < min_len:
                min_len = dp[(1 << n) - 1][i]
                last = i
        
        # Reconstruct the superstring from parent pointers
        mask = (1 << n) - 1
        s = []
        while mask:
            prev = parent[mask][last]
            if prev == -1:
                s.append(words[last])
            else:
                # Add the non-overlapping part of last
                s.append(words[last][overlap[prev][last]:])
            mask ^= (1 << last)
            last = prev
        
        # Since we added the strings in reverse order, reverse the list
        return ''.join(s[::-1])

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

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