HackerRank Merge the Tools! solution in python | python problem solution

• A string, , of length  where .
• An integer, , where  is a factor of .

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

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

Given  and , print  lines where each line  denotes string .

Example

There are three substrings of length  to consider: ‘AAA’, ‘BCA’ and ‘DDE’. The first substring is all ‘A’ characters, so . The second substring has all distinct characters, so . The third substring has  different characters, so . Note that a subsequence maintains the original order of characters encountered. The order of characters in each subsequence shown is important.

Function Description

Complete the merge_the_tools function in the editor below.

merge_the_tools has the following parameters:

• string s: the string to analyze
• int k: the size of substrings to analyze

Prints

Print each subsequence on a new line. There will be  of them. No return value is expected.

The first line contains a single string, .
The second line contains an integer, , the length of each substring.

• , where  is the length of 
• It is guaranteed that  is a multiple of .

STDIN       Function
-----       --------
AABCAAADA   s = 'AABCAAADA'
3           k = 3

AB
CA
AD

Split  into  equal parts of length . Convert each  to  by removing any subsequent occurrences of non-distinct characters in :

Print each  on a new line.

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))