Maventic Technical Question

Technical Round Question (Pattern in C/C++)

In one of the recent interviews (Technical Round 2) of Maventic, the following design pattern was asked

In this article, I'm going to explain, how we can easily achieve this design pattern in C programming language

If you are familiar with the programming languages. Then this is very easy to understand this concept.

nowadays many companies asking this concept in the Technical rounds, so we need to clear this concept to get the place.

Question: Write a program to design the pattern based on n 

          

example : n = 9

output:

Solution

This is very easy to design, first of all, we need to analyse the pattern. This is the combination of * and blank space.

we only need to identify in which position we need to print * and don't worry about the blank spaces.

 [ "-" represent a blank space][ and n = 9]

Here we can solve this in some steps

 (i) this program is the shape of a matrix. so we need to use two for loops to solve this problem.
   
     example 1: 

        for(i=1;i<=n;i++)
           {
             for(j=1;j<=n;j++)
              {
                  if(condition)
                         printf(" *");
                        else
                         printf("  ");//blank space
                   
              }
             printf("\n");
           }

(ii) ith 1st and n(9)th index jth 1st and n(9)th index as we can see here, each position we need to print '*'.
    in the side loop condition, we can write it as below,
 
    example 2:

      if(i==1 || i==n || j==1 || j==n)
         printf(" *");
       else
        printf("  ");

    then output comes like the box.

   


 (iii) now we need to draw like:

   
      here we can see it's printing * where-where ith index and
      jth index are equals except middle mean [(n/2)+1=5]
      so we can write code for this:
 
     example 3 :

        if(i==j && j!=(n/2)+1)
         printf(" *");
       else
         printf("  ");

 (iv) Now last pattern, reverse of step(iii) pattern
      

   
     so here we can analyse it's decreasing order of jth index
     9,8,7,6,4,3,2,1 except middle means [n/2+1=5]
   
     For this we can write the condition:-

     example 4 :

       if(j==(n+1)-i && j!=(n/2)+1)
          printf(" *");
        else
          printf("  ");

In the above example, we declared some variation of pattern based on condition.

Now we combined example3 and example4 to achieve like below

   

Now needs to add the condition as well 
example 5 :
       if(i==j && j!=(n/2)+1 || j==(n+1)-i && j!=(n/2)+1)
          printf(" *");
        else
          printf("  ");

And finally, If we combined example 5 with example 2 then the output will be 

Final output:

             

Hence,  below the final code I am writing

#include<stdio.h>
#include<conio.h>

void main()
{
 int n;
 clrscr();
 printf("Enter num: ");
  scanf("%d",&n);

 for( i=1;i<=n;i++)
 {
for(int j=1;j<=n;j++)
 {

if(i==1 || j==1 || i==n|| j==n || i==j && j!=(n/2)+1 || j==((n+1)-i) && j!=n/2+1)
{
     printf(" *");
  }
 else
      printf("  ");
}
 printf("\n");
 }

 getch();
}

below is some output I have attached. 

I hope this has cleared your concept, if yes please add your valuable comments, like subscribe and share with your friends. Thanks