查找不包含特定字符串的所有组合

Question

Is there any way to find all combinations of an array that not contains specific substrings? For example a=['a','b','c'] aaa aab aac aba abb ... ccc

but i don't want the substring ab so aaa aac aca acb ... ccc I use the code bellow but for combination of 20 chars and comb of 13 it takes too much time

import itertools

lista=[]
def foo(l):
    yield from itertools.product(*([l] * 3)) 

non=["ab"]

for x in foo('abc'):
    x=(''.join(x))
    for j in non:
        if j in x:
            value=1
            break
        else:
            value=0
    if (value==0):
        lista.append(x)

Answer 1

Instead of generating all strings then rejecting the ones containing any forbidden substrings, it is (asymptotically) more efficient to build the strings by backtracking, and reject any partial string which already contains a forbidden substring. We only need to test if the current partial string ends with any forbidden substring, which is faster than testing if it contains one.

Here's an implementation using a recursive generator function:

def strings_without(alphabet, k, avoid):
    def helper(s):
        if any(s.endswith(t) for t in avoid):
            pass
        elif len(s) == k:
            yield s
        else:
            for c in alphabet:
                yield from helper(s + c)
    return helper('')

Example:

>>> for s in strings_without('abc', 3, ['ab']):
...     print(s)
... 
aaa
aac
aca
acb
acc
baa
bac
bba
bbb
bbc
bca
bcb
bcc
caa
cac
cba
cbb
cbc
cca
ccb
ccc

For strings of length 13 from an alphabet of size 20, this should be a big improvement, but 20 ¹³ is an enormous number. So unless you are forbidding a lot of substrings, the number of solutions will be very large. No algorithm can possibly generate h strings of length k in less than O( hk ) time, so the running time for any efficient algorithm will still be output-sensitive .