CodeVita Round 1 2020 question Season 9 # Constellation

by - UniqueVerse1 on - 6:59 AM

Three characters { #, *, . } represents a constellation of stars and galaxies in space. Each galaxy is demarcated by # characters. There can be one or many stars in a given galaxy. Stars can only be in shape of vowels { A, E, I, O, U } . A collection of * in the shape of the vowels is a star. A star is contained in a 3x3 block. Stars cannot be overlapping. The dot(.) character denotes empty space.

Given 3xN matrix comprising of { #, *, . } character, find the galaxy and stars within them.

Note: Please pay attention to how vowel A is denoted in a 3x3 block in the examples section below.

- Constraints

3 <= N <= 10^5

- Input Format

Input consists of single integer N denoting number of columns.

- Output

Output contains vowels (stars) in order of their occurrence within the given galaxy. Galaxy itself is represented by # character.

- Timeout

- Explanation

Example 1

Input

18

* . * # * * * # * * * # * * * . * .

* . * # * . * # . * . # * * * * * *

* * * # * * * # * * * # * * * * . *

Output

U#O#I#EA

Explanation

As it can be seen that the stars make the image of the alphabets U, O, I, E and A respectively.

CodeVita Round 1 2020 question Season 9 # Constellation

Example 2

Input

12

* . * # . * * * # . * .

* . * # . . * . # * * *

* * * # . * * * # * . *

Output

U#I#A

Explanation

As it can be seen that the stars make the image of the alphabet U, I and A.

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Feel free to Provide your code in the comment section. We will be more than happy to see you enthusiasm form your side to crack it . Happy learning.

1 Comments

UnknownAugust 12, 2020 at 10:54 PM
def constellation(arr,n):
A=[['.','*','.'],
['*','*','*'],
['*','.','*']]

E=[['*','*','*'],
['*','*','*'],
['*','*','*']]

I=[['*','*','*'],
['.','*','.'],
['*','*','*']]

O=[['*','*','*'],
['*','.','*'],
['*','*','*']]

U=[['*','.','*'],
['*','.','*'],
['*','*','*']]

hash1=[['#','#','#'],
['#','#','#'],
['#','#','#']]
ans=[]
i=0

while(i+2<n):
aux_mat=[[arr[0][i],arr[0][i+1],arr[0][i+2]],
[arr[1][i],arr[1][i+1],arr[1][i+2]],
[arr[2][i],arr[2][i+1],arr[2][i+2]]]
if(aux_mat==A):
ans.append('A')
i+=3
elif(aux_mat==E):
ans.append('E')
i+=3
elif(aux_mat==I):
ans.append('I')
i+=3
elif(aux_mat==O):
ans.append('O')
i+=3
elif(aux_mat==U):
ans.append('U')
i+=3
elif((arr[0][i]=='#' and arr[1][i]=='#' and arr[2][i]=='#')):
#ans.append('#')
k=i
while((arr[0][k]=='#' and arr[1][k]=='#' and arr[2][k]=='#')):
#print(k)
ans.append('#')
k+=1
if(k==n):
break

i=k

#elif(i==9):pass

else:
i+=1
if(n-i==2 or n-i==1):
if((arr[0][i]=='#' and arr[1][i]=='#' and arr[2][i]=='#')):
k=i
while((arr[0][k]=='#' and arr[1][k]=='#' and arr[2][k]=='#')):
#print(k)
ans.append('#')
k+=1
if(k==n):
break

return ans

N=int(input())
v=[]
arr=[]
for i in range(3):
holder=list(str(j) for j in input().strip().split(' '))
arr.append(holder)
holder=[]
constellation(arr,N)

CodeVita Round 1 2020 question Season 9 # Constellation

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