HackerRank Merge the Tools! solution in python

In this HackerRank Merge the tools problem solution in python Consider the following:

• A string, s, of length n where s = c0c1...cn-1.
• An integer, k, where k is a factor of n.

We can split s into n/k substrings where each subtring, ti, consists of a contiguous block of k characters in s. Then, use each ti to create string ui such that:

• The characters in ui are a subsequence of the characters in ti.
• Any repeat occurrence of a character is removed from the string such that each character in ui occurs exactly once. In other words, if the character at some index j in ti occurs at a previous index <j in ti, then do not include the character in string ui.

Given s and k, print n/k lines where each line i denotes string ui.

Problem solution in Python 2 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
s = raw_input().strip()
k = int(raw_input())
i = 0
while i < len(s):
    a = s[i:i+k]
    output = ""
    for x in a:
        if x not in output:
            output += x
    print output
    i += k

Problem solution in Python 3 programming.

def merge_the_tools(string, k):
    for part in zip(*[iter(string)] * k):
        d = dict()
        print(''.join([ d.setdefault(c, c) for c in part if c not in d ]))

Problem solution in pypy programming.

def merge_the_tools(string, k):
    # your code goes here
    for x in xrange(0,len(string),k):
        u_list=list(set(string[x:x+k]))
        print ''.join(u_list)

Problem solution in pypy3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
s=input()
k=int(input())
ln=len(s)

for i in range(0,ln,k):
    ss=s[i:i+k]
    sss=[]
    for x in ss:
        if x not in sss:
            sss.append(x)

    print (''.join(sss))