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Problem solution · Python

Find the Shortest Superstring

Find the Shortest Superstring: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
49 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Find the Shortest Superstring, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 49 lines of Python from the credited upstream file find-the-shortest-superstring.py.
  • The implementation visibly relies on sequence storage, ordered lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Shortest Superstring · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n^2 * (l^2 + 2^n))# Space: O(n^2) class Solution(object):    def shortestSuperstring(self, A):        """        :type A: List[str]        :rtype: str        """        n = len(A)        overlaps = [[0]*n for _ in xrange(n)]        for i, x in enumerate(A):            for j, y in enumerate(A):                for l in reversed(xrange(min(len(x), len(y)))):                    if y[:l].startswith(x[len(x)-l:]):                        overlaps[i][j] = l                        break         dp = [[0]*n for _ in xrange(1<<n)]        prev = [[None]*n for _ in xrange(1<<n)]        for mask in xrange(1, 1<<n):            for bit in xrange(n):                if ((mask>>bit) & 1) == 0:                    continue                prev_mask = mask^(1<<bit)                for i in xrange(n):                    if ((prev_mask>>i) & 1) == 0:                        continue                    value = dp[prev_mask][i] + overlaps[i][bit]                    if value > dp[mask][bit]:                        dp[mask][bit] = value                        prev[mask][bit] = i                bit = max(xrange(n), key = dp[-1].__getitem__)        words = []        mask = (1<<n)-1        while bit is not None:            words.append(bit)            mask, bit = mask^(1<<bit), prev[mask][bit]        words.reverse()        lookup = set(words)        words.extend([i for i in xrange(n) if i not in lookup])         result = [A[words[0]]]        for i in xrange(1, len(words)):            overlap = overlaps[words[i-1]][words[i]]            result.append(A[words[i]][overlap:])        return "".join(result)

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