# Hackerrank Combination With Replacement Solution

itertools.combinations_with_replacement(iterable, r)
This tool returns  length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

Combinations are emitted in lexicographic sorted order. So, if the input iterable is sorted, the combination tuples will be produced in sorted order.

Sample Code

>>> from itertools import combinations_with_replacement
>>>
>>> print list(combinations_with_replacement('12345',2))
[('1', '1'), ('1', '2'), ('1', '3'), ('1', '4'), ('1', '5'), ('2', '2'), ('2', '3'), ('2', '4'), ('2', '5'), ('3', '3'), ('3', '4'), ('3', '5'), ('4', '4'), ('4', '5'), ('5', '5')]
>>>
>>> A = [1,1,3,3,3]
>>> print list(combinations(A,2))
[(1, 1), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (1, 3), (3, 3), (3, 3), (3, 3)]


You are given a string .
Your task is to print all possible size  replacement combinations of the string in lexicographic sorted order.

Input Format

A single line containing the string  and integer value  separated by a space.

Constraints

The string contains only UPPERCASE characters.

Output Format

Print the combinations with their replacements of string  on separate lines.

Sample Input

HACK 2


Sample Output

AA
AC
AH
AK
CC
CH
CK
HH
HK
KK

### Solution in Python

from itertools import combinations_with_replacement
string, n = input().split()
data = ["".join(sorted(i)) for i in combinations_with_replacement(string,int(n))]
data.sort()
print("\n".join(data))