Yes, it does solve it. It produces the expected output (`a` should occur twice)

Here's solution that doesn't use itertools.groupby() https://docs.python.org/3/library/itertools.html?highlight=i... :

    result = []
    prev, count = None, 0
    for c in s+"\0":  # note: assuming "\0" can't be in `s`
        if c != prev: # starting new sequence
           if count:  # save count for old sequence
              result.append((prev, count))
           prev, count = c, 1  # reset count
        else: # continuation of old sequence
           count += 1