Python: Sub sequence search

the absolute most strait forward way I can think of is using itertools.combinations

import itertools

sequence = "abccabac"

subsequence = "abc"

for combo in itertools.combinations(enumerate(sequence), len(subsequence)):
    if "".join(pair[1] for pair in combo) == subsequence:
        print([pair[0] for pair in combo])

If your actual sequence contains lots of characters that are not even in the subsequence then filtering out the irrelevant characters before starting the combinations would definitely provide a substantial efficiency boost:

char_generator = (pair for pair in enumerate(sequence) if pair[1] in subsequence)
for combo in itertools.combinations(char_generator, len(subsequence)):
    ...

as well instead of joining each combination you can just use all which will only check characters until one is found to be not equal:

char_generator = (pair for pair in enumerate(sequence) if pair[1] in subsequence)
for combo in itertools.combinations(char_generator, len(subsequence)):
    if all(pair[1]==char for pair,char in zip(combo,subsequence)):
        print([pair[0] for pair in combo])

Here is a different alternative I just put together, I imagine the algorithm could be optimized even more but this is about the best I could come up with. The sub_sequence_generator creates a list of indices for each letter of the sub sequence, then the sub_helper for each level of the recursion goes through all indices for the next letter of the subsequence starting after the indice of the last letter.

import itertools
import bisect

def sub_helper(cur_i, combo, all_options):
    #cur_i is the index of the letter in the subsequence
    #cur_idx is the index of the sequence which contains that letter
    if cur_i>0:
        prev_i = combo[cur_i-1]
    else:
        prev_i = -1
    cur_options = all_options[cur_i]
    start_i = bisect.bisect(cur_options,prev_i)
    cur_i_gen = itertools.islice(cur_options,start_i,None)
    if cur_i+1 == len(all_options): #last one:
        for cur_idx in cur_i_gen:
            combo[cur_i] = cur_idx
            yield tuple(combo)
    else:
        for cur_idx in cur_i_gen:
            combo[cur_i] = cur_idx
            yield from sub_helper(cur_i+1, combo, all_options)


def sub_sequence_generator(sequence,sub_seq):
    indices_map = {c:[] for c in set(sub_seq)} #create a list for each character
    for i,c in enumerate(sequence):
        try:
            indices_map[c].append(i,) #the helper is simpler if they are all stored as single element tuples
        except KeyError:
            pass

    # now we have indices for each character of the sub_sequence
    # we just need to put them in a sequence that corelates to the subsequence
    chr_indices = tuple(indices_map[c] for c in sub_seq)

    return sub_helper(0,[None]*len(chr_indices), chr_indices)

sequence = "abccabac"
subsequence = "abc"

for idxs in sub_sequence_generator(sequence,subsequence):
    print(idxs)

In terms of memory this will make a list for each character of the subsequence, containing the indices in the main sequence where that character is present. A tuple of these lists, and a list of the indices that is continually updated and a tuple for each time a combination is yielded, and iterators like islice so this is extremely memory efficient, however since I have not extensively tested it I cannot give any guarantee it is bug free.