✅ One Line Solution
CodeTime complexity: O(Cnm​∗m). Space complexity: O(Cnm​∗m).Ref
# Code
Time complexity: $$O(C_n^m*m)$$. Space complexity: $$O(C_n^m*m)$$.
```python3
class CombinationIterator:
    def __init__(self, *q):
        self.next,self.hasNext=(b:=[*map(''.join,combinations(*q))][::-1]).pop,lambda:bool(b)
```
[Ref](https://leetcode.com/problems/iterator-for-combination/solutions/451385/3-lines-python-ruby/)
['String', 'Design', 'Iterator', 'Python', 'Python3']