# 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):
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))  ```