<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:
        # Traveling Salesman Problem in disguise (NP Complete = O(n^2 * 2^n) -> finding overlap of each word duo then finding the bitmask with recursion)
        # this a top-down DP using recursion solution, although you could also do it bottom up with a iteration
        # Idea: think of this as graph proble where we need to visit each node with min overall cost, words = nodes, and nonoverlapping substrings btwn two words are costs
        # i.e. if the word overlaps then the cost is only the non-overlap
        # this also uses bit manipulation to represent the states for efficiency

        n = len(words)
        overlap = [['']*n for _ in range(n)]

        def getOverlap(a, b):
            # given two words, find the overlap between the suffix of word 'a' and the prefix of word 'b', return the nonoverlap suffix of word 'b'
            l = 0
            i = 1
            while i <= min(len(a), len(b)):
                if a[len(a)-i:] == b[:i]:
                    l = i
                i +=1
            return b[l:]
        
        # calculate overlaps btwn every word combo (excluding self)
        for i in range(n):
            for j in range(n):
                if i != j:
                    overlap[i][j] = getOverlap(words[i], words[j])
        
        def TSP(prev, mask, dp):
            # prev represents the last word we chose, the mask represents bitmask of already used words, the dp is our recording of each state
            # the final answer should be the state with the least amount of letters

            if (prev, mask) in dp: return dp[(prev, mask)]  # memoized case
            if mask == ((1 << n) - 1): return '' # base case all words have been used
            ans = '*' * 1000000  # some very large string

            # Iterate through each word, if the word is already in our ans then we skip, else calculate the overlap with prev (if not first word);
            # lastly, caculate the final length of the ans by running tsp recursively and taking the min btwn current ans length OR new TSP+word length
            for i in range(n):
                if mask & (1<<i): continue  # & = bitwise and
                word = overlap[prev][i] if prev != -1 else words[i]
                ans = min(ans, word+TSP(i, mask | (1<<i), dp), key=len)  # | = bitwise or, and 'key=len' means min compares by length rather than value
            dp[(prev, mask)] = ans
            return ans
        return TSP(-1, 0, {})  # prev starts as -1 for null, mask starts at 0 for no words, and dict for dp


        
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