First of all, whenever you see this ... this is a big no-no:
'9'*work_dict[9] + '8'*work_dict[8] + '7'*work_dict[7] +
'6'*work_dict[6] + '5'*work_dict[5] + '4'*work_dict[4] +
'3'*work_dict[3] + '2'*work_dict[2] + '1'* work_dict[1] +
'0'*work_dict[0]
it could be replaced by a simple
''.join(str(i)'*work_dict[i] for i in reversed(range(10)))
# or if you want to use Python >=3.6 f-string notation:
''.join(f'{i}'*work_dict[i] for i in reversed(range(10)))
Of course, the initialization of the work_dict is similar. And in fact, you don't need to initialize it if you take care to use dict.get instead of dict[]:
work_dict[number] = work_dict[number] + 1
# is equivalent to
work_dict[number] = work_dict.get(number, 0) + 1 # default to 0 if not in dict
If you want to have it as compact and "functional" as possible, it would be much easier to just sort the entire input and output it:
def maximum_number(lst):
return int(''.join(reversed(sorted(''.join(str(x) for x in lst)))))
However, note that this doesn't work on the empty list (which might be okay, depending on the specification of the function).
It should also be mentioned that
- it is harder to write than the "manual loop" variant, which can be important in an interview
- it might be harder to read and thus to debug, but I believe that this is up to the eye of the beholder to determine