Hackerrank - Encryption Solution
- Published
- Updated
- Reading time
- 3 min read
- Length
- 530 words
An English text needs to be encrypted using the following encryption scheme.
First, the spaces are removed from the text. Let be the length of this text.
Then, characters are written into a grid, whose rows and columns have the following constraints:
For example, the sentence , after removing spaces is characters long. is between and , so it is written in the form of a grid with 7 rows and 8 columns.
ifmanwas
meanttos
tayonthe
groundgo
dwouldha
vegivenu
sroots- Ensure that
- If multiple grids satisfy the above conditions, choose the one with the minimum area, i.e. .
The encoded message is obtained by displaying the characters in a column, inserting a space, and then displaying the next column and inserting a space, and so on. For example, the encoded message for the above rectangle is:
imtgdvs fearwer mayoogo anouuio ntnnlvt wttddes aohghn sseoau
You will be given a message to encode and print.
Function Description
Complete the encryption function in the editor below. It should return a single string composed as described.
encryption has the following parameter(s):
- s: a string to encrypt
Input Format
One line of text, the string
Constraints
is comprised only of characters in the range ascii[a-z].
Output Format
Print the encoded message on one line as described.
Sample Input
haveanicedaySample Output 0
hae and via ecyExplanation 0
, is between and .
Rewritten with rows and columns:
have
anic
edaySample Input 1
feedthedog Sample Output 1
fto ehg ee ddExplanation 1
, is between and .
Rewritten with rows and columns:
feed
thed
ogSample Input 2
chilloutSample Output 2
clu hlt ioExplanation 2
, is between and .
Rewritten with columns and rows ( so we have to use .)
chi
llo
utSolution in Python#
import math
from collections import defaultdict
def encryption(text):
text = text.replace(" ","")
root= math.sqrt(len(text))
r = math.floor(root)
c = math.ceil(root)
d = defaultdict(str)
for i in range(0,len(text),c):
sub = text[i:i+c]
for x in range(len(sub)):
d[x]+=sub[x]
return(list(d.values()))
print(*encryption(input()))Explanation#
We remove any empty spaces from our text
text = text.replace(" ","")Then we use math.floor and math.ceil function to calculate the required number of rows and columns denoted by r and c respectively.
root= math.sqrt(len(text))
r = math.floor(root)
c = math.ceil(root)Using default dict we will initialize an empty dictionary.
d = defaultdict(str)We use for loop to divide our string into equal number of columns
for i in range(0,len(text),c):
sub = text[i:i+c]
print(sub)Example
>>> haveaniceday
have
anic
edayInstead of printing the value of our variable sub we add each letter to our default dict
Loop 1 >>> d = {0:"h", 1:"a", 2:"v", 3:"e"}
Loop 2 >>> d = {0:"ha", 1:"an", 2:"vi", 3:"ec"}
Loop 3 >>> d = {0:"hae", 1:"and", 2:"via", 3:"ecy"}>>> list(d.values())
['hae', 'and', 'via', 'ecy']Now the final step remaining is to print the values separated by a space
>>> d = list(d.values())
>>> print(*d)
hae and via ecy
>>> print(" ".join(d))
hae and via ecy