How to find the longest common suffix prefix between two strings in python in a pythonic way possibly using library functions?

Question

Let's say I have two strings, s1 = "1234" and s2 ="34567", so longest common suffix prefix between s1 and s2 is "34". I want to know if there exists any pythonic way to get this matching part ("34") real quick.

I can do it in a naive way like this one below, but I would love to know if there is an interesting library function or algorithm to get this done.

s1 = "1234"
s2 = "34567"
length1 = len(s1) 
length2 = len(s2)

length = (length1 if length1<= length2 else length2)

for i in reversed(range(0, length)):
    if s1[-i - 1:] == s2[:i + 1]:
        print(s1[-i - 1:])
        break
    elif i > 0:
        continue
    else:
        print("no common suffix prefix")

Output:

34

I want something compact and smart!

Answer 1

The logic in your algorithm is about as straightforward as you can get, but you can definitely compactify the notation. For example, checking a prefix of size n against a suffix of size n is simply:

s1[-n:] == s2[:n]

The ternary operator you use to check string lengths is

min(len(s1), len(s2))

A range can go backwards by itself. The reverse of range(x) is

range(x - 1, -1, -1)

You can create an iterator that checks this for each decreasing value of n and return the first non-zero result. Luckily, next accepts a second argument that represents the default if the iterator is empty:

common = next((s2[:n] for n in range(min(len(s1), len(s2)) - 1, -1, -1) if s1[-n:] == s2[:n]), '')

That's the obligatory one-liner. A more legible solution might be:

def common_fix(s1, s2):
    steps = range(min(len(s1), len(s2)) - 1, -1, -1)
    return next((s2[:n] for n in steps if s1[-n:] == s2[:n]), '')

As a rule, keep your functionally and printing separate. Get a value, then process it (whether by printing or something else)