itertools.permutations() in Python - HackerRank Solution
Problem
itertools.permutations(iterable[, r])
This tool returns successive r length permutations of elements in an iterable.
If r is not specified or is None, then r defaults to the length of the iterable, and all possible full length permutations are generated.
Permutations are printed in a lexicographic sorted order. So, if the input
iterable is sorted, the permutation tuples will be produced in a sorted order.
Sample Code :
>>> from itertools import permutations >>> print permutations(['1','2','3']) <itertools.permutations object at 0x02A45210> >>> >>> print list(permutations(['1','2','3'])) [('1', '2', '3'), ('1', '3', '2'), ('2', '1', '3'), ('2', '3', '1'), ('3', '1', '2'), ('3', '2', '1')] >>> >>> print list(permutations(['1','2','3'],2)) [('1', '2'), ('1', '3'), ('2', '1'), ('2', '3'), ('3', '1'), ('3', '2')] >>> >>> print list(permutations('abc',3)) [('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]
Task :
Your task is to print all possible permutations of size k of the string in
lexicographic sorted order.
Input Format :
A single line containing the space separated string S and the integer value k.
Constraints :
- 0 < k <= len(S)
Output Format :
Sample Input :
HACK 2
Sample Output :
AC AH AK CA CH CK HA HC HK KA KC KH
Solution :
1 2 3 4 5 6 7 8 9 10 11 12 | # itertools.permutations() in Python - Hacker Rank Solution # Python 3 # itertools.permutations() in Python - Hacker Rank Solution START from itertools import permutations s,k = input().split() words = list(permutations(s,int(k))) words = sorted(words, reverse=False) for word in words: print(*word,sep='') # itertools.permutations() in Python - Hacker Rank Solution END |
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Here my solution:
from itertools import combinations, combinations_with_replacement
s, n = input().split()
for i in combinations_with_replacement(sorted(s), int(n)):
print("".join(i))