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

Difficulty: introductory

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)    # make a big string
    return "".join(c for c in "0123456789" if c not in s)

import string

def unused_digits(*args):
  #args is tuple of numbers
  #for each number we find, lets turn it into a string
  #then for each char in num str just add it to a set of found digits
  found_digits_chars = {char for n in args for char in str(n)}

  #let's say i have set of all digits {'0' to '9'}
  all_digits_chars = set(string.digits)
  #then just subtract found digits from set of all digits
  unused_digits_chars = all_digits_chars - found_digits_chars

  #then of course change format of answer to ordered strings
  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))))