Problem
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.
Example
s = ‘AAABCADDE’
k = 3
There are three substrings of length 3 to consider: ‘AAA’, ‘BCA’ and ‘DDE’. The first substring is all ‘A’ characters, so u1 = ‘A’. The second substring has all distinct characters, so u2 = ‘BCA’. The third substring has 2 different characters, so u3 = ‘DE’. 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 n/k of them. No return value is expected.
Input Format
The first line contains a single string, s.
The second line contains an integer, k, the length of each substring.
Constraints
- 1 ≤ n ≤ 104
- 1 ≤ k ≤ n
- It is guaranteed that n is a multiple of k.
Sample Input
STDIN Function----- --------AABCAAADA s = 'AABCAAADA'3 k = 3
Sample Output
ABCAAD
Explanation
Split s into n/k = 9/3 = 3 equal parts of length k = 3. Convert each ti to ui by removing any subsequent occurrences of non-distinct characters in ti:
- t0 = “AAB” -> u0 = “AB”
- t1 = “CAA” -> u1 = “CA”
- t2 = “ADA” -> u2 = “AD”
Print each ui on a new line.
Solution to HackerRank Merge the Tools! In Python
import textwrapdef merge_the_tools(s, k): # your code goes here for i in textwrap.wrap(s, k): l = list() for j in i: if not j in l: l.extend(j) print(''.join(l))if __name__ == '__main__': string, k = input(), int(input()) merge_the_tools(string, k)
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