CONSTPROB Editorial

editorial
1

Problem Explanation

We are given a N \times 3 matrix of characters {#, *, .}. The * form vowels {a, e, i, o, u}, ‘.’ denotes empty spaces and # denotes a partition between galaxies.

Approach

we can iterate a matrix till n-2 and check for the 3 \times 3 sub - matrix if it’s characters match with the composition of any vowels. If we encounter #, we can directly output # in the ans string.

Code

#include<bits/stdc++.h>
using namespace std;
 
int main()
{
    int n;cin>>n;
    char co[3][n];
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<n;j++)
        cin>>co[i][j];
    }
    for(int i=0;i<n-2;i++)
    {
        if(co[0][i]=='#') {cout<<"#";continue;}
        if(co[0][i]=='.' && co[0][i+1]=='*' && co[0][i+2]=='.')
        {
            if(co[1][i]=='*' && co[1][i+1]=='*' && co[1][i+2]=='*')
            if(co[2][i]=='*' and co[2][i+1]=='.' and co[2][i+2]=='*')
            cout<<"A";i+=2;continue;
        }
        if(co[0][i]=='*' and co[0][i+1]=='*' and co[0][i+2]=='*')
        {
            if (co[1][i]=='*' and co[1][i+1]=='*' and co[1][i+2]=='*')
            {
                if (co[2][i]=='*' and co[2][i+1]=='*' and co[2][i+2]=='*')
                {cout<<"E";i+=2;continue;}
            }
            else if(co[1][i]=='.' and co[1][i+1]=='*' and co[1][i+2]=='.')
            {
                 if (co[2][i]=='*' and co[2][i+1]=='*' and co[2][i+2]=='*')
                 {cout<<"I";i+=2;continue;}
            }
            else if(co[1][i]=='*' and co[1][i+1]=='.' and co[1][i+2]=='*')
            {
                if(co[2][i]=='*' and co[2][i+1]=='*' and co[2][i+2]=='*')
                {cout<<"O";i+=2;continue;}
            }
        }
        if(co[0][i]=='*' and co[0][i+1]=='.' and co[0][i+2]=='*')
        if(co[1][i]=='*' and co[1][i+1]=='.' and co[1][i+2]=='*')
        if(co[2][i]=='*' and co[2][i+1]=='*' and co[2][i+2]=='*')
        {cout<<"U";i+=2;continue;}
    }
}