Given few numbers, you need to print out the digits that are not being used.

Example:

```python
unused_digits(12, 34, 56, 78) # "09"
unused_digits(2015, 8, 26) # "3479"
```

Note:

- Result string should be sorted
- The test case won't pass Integer with leading zero

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Starter code: def unused_digits(*args):
	

Solutions:
def unused_digits(*args):
	return ''.join((number for number in '0123456789' if number not in str(args)))


def unused_digits(*l):
	digits_set = set(list('0123456789'))
	test_set = set(''.join((str(i) for i in l)))
	d = digits_set.difference(test_set)
	r = ''.join(sorted(d))
	return r


def unused_digits(*args):
	s = ''.join(map(str, args))
	return ''.join((str(i) for i in range(0, 10) if str(i) not in s))


def unused_digits(*args):
	return ''.join(sorted(set('0123456789') - set(str(args))))


def unused_digits(*ints):
	return ''.join(sorted(list(set('0123456789') - set((c for i in ints for c in str(i))))))


def unused_digits(*args):
	s = ''.join((str(a) for a in args))
	return ''.join((c for c in '0123456789' if c not in s))


import string

def unused_digits(*args):
	found_digits_chars = {char for n in args for char in str(n)}
	all_digits_chars = set(string.digits)
	unused_digits_chars = all_digits_chars - found_digits_chars
	return ''.join(sorted(list(unused_digits_chars)))


def unused_digits(*args):
	return ''.join(sorted(set('0123456789') - {c for s in map(str, args) for c in s}))


def unused_digits(*args):
	return ''.join(sorted(set('1234567890') - set(''.join((str(x) for x in args)))))