HackerRank Merge the Tools! solution in python | python problem solution
Table of Contents
Consider the following:
- 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.
Input Format
The first line contains a single string, .
The second line contains an integer, , the length of each substring.
Constraints
- , where is the length of
- It is guaranteed that is a multiple of .
Sample Input
STDIN Function
----- --------
AABCAAADA s = 'AABCAAADA'
3 k = 3
Sample Output
AB
CA
AD
Explanation
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))