Merge the Tools! Solution | HackerRank | Python

The Problem

Consider the following:

• n is a length of a string s.
• k is a factor of n.

We can split s into $\frac n k$ substrings where each subtring, $t_i$ , consists of a contiguous block of k characters in s. Then, use each $t_i$ to create string $u_i$ such that:

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

Given s and k, print $\frac n k$ lines where each line i denotes string $u_i$ .

Print each subsequence on a new line. There will be $\frac n k$ of them. No return value is expected.

The Input

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

Constraints

• $1 \le n \le 10^4$ , where n is the length of s
• $1 \le k \le n$
• It is guaranteed that n is a multiple of k.

Sample:

AABCAAADA
3

The Output

Sample:

AB
CA
AD

Explanation

Split s into $\frac n k = \frac 9 3 = 3$ equal parts of length k = 3. Convert each $t_i$ to $u_i$ by removing any subsequent occurrences of non-distinct characters in $t_i$ :

1. $t_0 = \text{\textquotedblleft} AAB \text{\textquotedblright} \rightarrow u_0 = \text{\textquotedblleft} AB \text{\textquotedblright}$
2. $t_1 = \text{\textquotedblleft} CAA \text{\textquotedblright} \rightarrow u_1 = \text{\textquotedblleft} CA \text{\textquotedblright}$
3. $t_2 = \text{\textquotedblleft} ADA \text{\textquotedblright} \rightarrow u_2 = \text{\textquotedblleft} AD \text{\textquotedblright}$

Print each $u_i$ on a new line.

The Solution

The Code

Source Code 1

def merge_the_tools(string, k):
    l = len(string)//k
    for i in range(l):
        print(''.join(dict.fromkeys(string[i*k:(i*k)+k])))

Source Code 2

from collections import OrderedDict as od

def merge_the_tools(string, k):
    l = len(string)//k
    for i in range(l):
        print(''.join(od.fromkeys(string[i*k:(i*k)+k])))

Original Source

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