find longest common substring in space O(n)

PEP 8

You are following most of the PEP 8 style guidelines. One that you are breaking is method names should be snake_case; your function should be named longest_common_substring.

Tricky Code

Your dp matrix is properly allocated to the size m+1 by n+1.

When you index your matrix, you access [i-1][j-1] with \$0 \le i \lt m\$ and \$0 \le j \lt n\$. This means you never access the last allocated row m or the last allocated column n, but instead rely on accessing the -1 row and the -1 column wrapping around to reach these "unused" spaces. This is "surprising" code at best, and "crashing" code if translated to a different programming language.

It would be better to add one to the indices used to index the dp matrix. The simplest way would be to start the i and j enumerations at one:

    for i, ic in enumerate(s1, 1):
        for j, jc in enumerate(s2, 1):

Useless else

Expand out this ... if ... else ... statement:

            max_len = dp[i][j] if len(max_len) < len(dp[i][j]) else max_len

Initially, this produces:

            if len(max_len) < len(dp[i][j]):
                max_len = dp[i][j]
            else:
                max_len = max_len

But we can immediately see the else: clause is a no-op, and can be removed:

            if len(max_len) < len(dp[i][j]):
                max_len = dp[i][j]

Which reads much more clearly than the original.

From \$O(n m)\$ to \$O(n)\$ space

During the first iteration of outer loop, you only access rows -1 and 0. During the second iteration of outer loop, you only access rows 0 and 1. During the third iteration of outer loop, you only access rows 1 and 2. Etc. You only need two rows of the dp matrix!

More over, you create the 0 row from the -1 row, you create the 1 from the 0 row, you create the 2 row from the 1 row, and so on.

Do you really need to keep the dp matrix? Or could you use a previous_row and a current_row? Only storing two length n rows reduces your space to \$O(2n)\$, which is simply \$O(n)\$.

You don't specify what n is, so I'm going to assume that n is the total combined length of the two strings s1 and s2 that are input.

Imagine you have a hashing function h(s) that computes a fixed-size hash value for a given string s. That is, given a string s = "abc" the function h(s) will return some number X that is computed from s, so that changing the contents of s will change the returned value X.

If you keep an array of hash values for the substrings of a given length k for each starting position i, you will have a total size of sizeof(h(s1)) * (n - 2k + 2) which is easily \$O(n)\$.

So, start from the longest possible substring and work your way down, computing the hashes of all substrings then checking to see if the two strings share a common substring. If two hashes are equal you can compare the resulting substrings and return if you find a match. If you get to k==1 and find no match, then there's no match to be had.

With that outline, I'll point out that python provides a built-in called hash() that does what you want.