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/

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=itertools#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