<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>

class Solution:
    def shortestSuperstring(self, words: List[str]) -> str:

        def connect(i, j):
            word1, word2 = words[i], words[j]
            common = []

            for k in range(len(word1) + 1):
                if k == 0 or word1[-k:] == word2[:k]:
                    common.append((word2[k:], len(word2) - k))

            return common[-1]

        n = len(words)
        mask = 1 << n

        
        dp = [[(float('inf'), -1)] * n for _ in range(mask)]

        for i in range(n):
            dp[1 << i][i] = (len(words[i]), -1)

        for curr_mask in range(mask):
            used_index = [i for i in range(n) if curr_mask & (1 << i)] 

            for i, j in permutations(used_index, 2):
                dp[curr_mask][j] = min(dp[curr_mask][j], (dp[curr_mask ^ (1 << j)][i][0] + connect(i, j)[1], i)) 

        curr_element = min(dp[-1])
        shotest_len, prev_index = min(dp[-1])
        
        curr_index = dp[-1].index(curr_element)
        curr_mask = mask - 1
        res = ''

        while prev_index != -1:
            res = connect(prev_index, curr_index)[0] + res
            curr_mask -= 1 << curr_index
            prev_index, curr_index = dp[curr_mask][prev_index][1], prev_index

        return words[curr_index] + res


        
        
19000