I've written a short python script that attempts to solve the problem of finding the longest common substring using the dynamic programming technique. It is meant to be generalised so I could plug in any number of strings and it would find the longest common substring.

def longest_common_substring(*strings):
    table = defaultdict(int)
    for pos in product(*(range(len(s)) for s in strings)):
        same = len(set(s[i] for s, i in zip(strings, pos))) is 1
        table[pos] = table[tuple(map(lambda n: n - 1, pos))] + 1 if same else 0
    return max(table.items(), key=operator.itemgetter(1))

This works fine for a small number of short strings, but the space and time complexity absolutely blows up with longer strings.

I got this from wikipedia, and since this approach is clearly terrible for multiple longer strings (or maybe my implementation is just bad!?), I am wondering what I could do to improve it? Wikipedia also metions a generalised suffix trees... I am not familiar with them at all so would that be a better approach?

Also, if its my implementation, I'd love to know what's wrong and what I could do better in terms of space complexity.


1 Answer 1


Sorry I've not examined too closely your code to make a comment, but just considering the problem as stated and the fact you are using Py3, I would probably solve it with itertools.accumulate, e.g.:

>>> import itertools as it
>>> import operator as op
>>> ss = ["thisishello", "dfdsishdllo", "ashsisdsdsf"]
>>> i, l = max(enumerate(it.accumulate(it.chain([0], zip(*ss)),
...                      lambda x, y: (x+1)*(len(set(y)) == 1))), key=op.itemgetter(1))
>>> i, l, ss[0][i-l:i]
(6, 3, 'sis')

Should work well for an arbitrary number of strings as it uses generators and doesn't create any intermediate data structures.
It does use the fact that a False equates to 0 with len(set(y)) == 1 but if that is uncomfortable you could simply replace with 1 if len(set(y)) == 1 else 0.

Note: I still wish that itertools.accumulate had an initial value argument much like functools.reduce has, would avoid the need of chaining the initial value to the iterable.

  • \$\begingroup\$ This is a very interesting approach, I've never really used accumulate much. Very interesting use case :) I'll be sure to test it out! \$\endgroup\$
    – Pavlin
    Commented Oct 29, 2016 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.