Constellation | TCS CodeVita 9 Solution ( Zone 1 ) 2020 | By CodingHumans |

Constellation

Problem Description

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

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.

Time Limit

1

Examples

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.




 
                                                                                                                                                                                                                                                                                                                                       
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.

Recommended: Please try your approach on your integrated development environment (IDE) first, before moving on to the solution.

Few words from CodingHumans : Don't Just copy paste the solution, try to analyze the problem and solve it without looking by taking the the solution as a hint or a reference . Your understanding of the solution matters.

HAVE A GOOD DAY 😁

 Solution

( C ++ )

#include<bits/stdc++.h>

using namespace std;

#define FIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

int main()

{

  FIO;

  int n,i=0; cin>>n;

  char a[3][n];

  for(int i=0;i<3;i++)

    for(int j=0;j<n;j++)

      cin>>a[i][j];

  while(i<n)

  {

    if(a[0][i]=='#'&&a[1][i]=='#'&&a[2][i]=='#')

    {

      cout<<"#", i++;

      continue;

    }

    else if(a[0][i]=='.'&&a[1][i]=='.'&&a[2][i]=='.')

    {

      i++;

      continue;

    }

   else if(a[0][i]=='.'&&a[0][i+1]=='*'&&a[0][i+2]=='.'&&a[1][i]=='*'&&a[1][i+1]=='*'&&a[1][i+2]=='*'&&a[2][i]=='*'&&a[2][i+1]=='.'&&a[2][i+2]=='*')

   {

     cout<<"A",i+=3;

     continue;

   }

   else if(a[0][i]=='*'&&a[0][i+1]=='*'&&a[0][i+2]=='*'&&a[1][i]=='*'&&a[1][i+1]=='*'&&a[1][i+2]=='*'&&a[2][i]=='*'&&a[2][i+1]=='*'&&a[2][i+2]=='*')

   {

     cout<<"E",i+=3;

     continue;

   }

   else if(a[0][i]=='*'&&a[0][i+1]=='*'&&a[0][i+2]=='*'&&a[1][i]=='.'&&a[1][i+1]=='*'&&a[1][i+2]=='.'&&a[2][i]=='*'&&a[2][i+1]=='*'&&a[2][i+2]=='*')

   {

     cout<<"I",i+=3;

     continue;

   }

   else if(a[0][i]=='*'&&a[0][i+1]=='*'&&a[0][i+2]=='*'&&a[1][i]=='*'&&a[1][i+1]=='.'&&a[1][i+2]=='*'&&a[2][i]=='*'&&a[2][i+1]=='*'&&a[2][i+2]=='*')

   {

     cout<<"O",i+=3;

     continue;

   }

   else if(a[0][i]=='*'&&a[0][i+1]=='.'&&a[0][i+2]=='*'&&a[1][i]=='*'&&a[1][i+1]=='.'&&a[1][i+2]=='*'&&a[2][i]=='*'&&a[2][i+1]=='*'&&a[2][i+2]=='*')

   {

     cout<<"U",i+=3;

     continue;

   }

   else

     i++;

  }

  return 0;

}


 Solution
( Python )

n=int(input())
li=[]
for i in range(3):
li.append(list(input().split()))
for i in range(n-2):
if li[0][i]=='#':
print("#",end="")
continue
if li[0][i]=='.' and li[0][i+1]=='*' and li[0][i+2]=='.':
if li[1][i]=='*' and li[1][i+1]=='*' and li[1][i+2]=='*':
if li[2][i]=='*' and li[2][i+1]=='.' and li[2][i+2]=='*':
print('A',end="")
if li[0][i]=='*' and li[0][i+1]=='*' and li[0][i+2]=='*':
if li[1][i]=='*' and li[1][i+1]=='*' and li[1][i+2]=='*':
if li[2][i]=='*' and li[2][i+1]=='*' and li[2][i+2]=='*':
print('E',end="")
if li[0][i]=='*' and li[0][i+1]=='*' and li[0][i+2]=='*':
if li[1][i]=='.' and li[1][i+1]=='*' and li[1][i+2]=='.':
if li[2][i]=='*' and li[2][i+1]=='*' and li[2][i+2]=='*':
print('I',end="")
if li[0][i]=='*' and li[0][i+1]=='*' and li[0][i+2]=='*':
if li[1][i]=='*' and li[1][i+1]=='.' and li[1][i+2]=='*':
if li[2][i]=='*' and li[2][i+1]=='*' and li[2][i+2]=='*':
print('O',end="")
if li[0][i]=='*' and li[0][i+1]=='.' and li[0][i+2]=='*':
if li[1][i]=='*' and li[1][i+1]=='.' and li[1][i+2]=='*':
if li[2][i]=='*' and li[2][i+1]=='*' and li[2][i+2]=='*':

print('U',end="")


If you have any doubts regarding this problem or  need the solution in other programming languages then leave a comment down below .