Understanding isomorphic strings algorithm

Question

I am understanding the following code to find if the strings are isomorphic or not. The code uses two hashes s_dict and t_dict respectively. I am assum

Answer 1

This was kind of fun to look at. Just for kicks, here's my solution using itertools.groupby

from itertools import groupby
from collections import defaultdict

def get_idx_count(word):
    """Turns a word into a series of tuples (idx, consecutive_chars)

    "aabbccaa" -> [[(0, 2), (3, 2)], [(1, 2)], [(2, 2)]]
    """
    lst = defaultdict(list)
    for idx, (grp, val) in enumerate(groupby(word)):
        lst[grp].append((idx, sum(1 for _ in val)))
    return sorted(list(lst.values()))

def is_isomorphic(a, b):
    return get_idx_count(a) == get_idx_count(b)

is_isomorphic('aabbcc', 'bbddcc')  # True
is_isomorphic('aabbaa', 'bbddcc')  # False

Rather than building the lists, I feel like I could do something more like:

from itertools import groupby
from collections import defaultdict

def is_isomorphic(a, b):
    a_idxs, b_idxs = defaultdict(set), defaultdict(set)
    for idx, ((a_grp, a_vals), (b_grp, b_vals)) in enumerate(zip(groupby(a), groupby(b))):
        if sum(1 for _ in a_vals) != sum(1 for _ in b_vals):
            return False
            # ensure sequence is of same length
        if a_grp in a_idxs and b_idxs[b_grp] != a_idxs[a_grp] or\
           b_grp in b_idxs and a_idxs[a_grp] != b_idxs[b_grp]:
            return False
            # ensure previous occurrences are matching groups
        a_idxs[a_grp].add(idx)
        b_idxs[b_grp].add(idx)
        # save indexes for future checks
    return True

But I haven't had a chance to test that whatsoever. It does, however, have the likelihood of exiting early, rather than having to build all the lists and comparing them at the end. It also skips a few sets of sorting that shouldn't be necessary anyway, except that we're pulling from dicts, so bleh. Pretty sure list.sort is faster than saving to to a collections.OrderedDict. I think.