Printing Pattern using Loops|||| hackerrank || C language || programming_info

In this problem, you need to print the pattern of the following form containing the numbers from  to .

                            4 4 4 4 4 4 4  
                            4 3 3 3 3 3 4   
                            4 3 2 2 2 3 4   
                            4 3 2 1 2 3 4   
                            4 3 2 2 2 3 4   
                            4 3 3 3 3 3 4   
                            4 4 4 4 4 4 4   

Input Format

The input will contain a single integer .

Constraints

Output Format

Print the pattern mentioned in the problem statement.

Sample Input 0

2

Sample Output 0

2 2 2
2 1 2
2 2 2

Sample Input 1

5

Sample Output 1

5 5 5 5 5 5 5 5 5 
5 4 4 4 4 4 4 4 5 
5 4 3 3 3 3 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 2 1 2 3 4 5 
5 4 3 2 2 2 3 4 5 
5 4 3 3 3 3 3 4 5 
5 4 4 4 4 4 4 4 5 
5 5 5 5 5 5 5 5 5

Sample Input 2

7

Sample Output 2

7 7 7 7 7 7 7 7 7 7 7 7 7 
7 6 6 6 6 6 6 6 6 6 6 6 7 
7 6 5 5 5 5 5 5 5 5 5 6 7 
7 6 5 4 4 4 4 4 4 4 5 6 7 
7 6 5 4 3 3 3 3 3 4 5 6 7 
7 6 5 4 3 2 2 2 3 4 5 6 7 
7 6 5 4 3 2 1 2 3 4 5 6 7 
7 6 5 4 3 2 2 2 3 4 5 6 7 
7 6 5 4 3 3 3 3 3 4 5 6 7 
7 6 5 4 4 4 4 4 4 4 5 6 7 
7 6 5 5 5 5 5 5 5 5 5 6 7 
7 6 6 6 6 6 6 6 6 6 6 6 7 
7 7 7 7 7 7 7 7 7 7 7 7 7 

Now here's the solution in C:-

code snippet:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() 
{
    int n,i,j,k;
    int s=0,e;
    scanf("%d", &n);
    k=(2*n)-1;
    int a[k][k];
    e=k-1;
    while(n!=0)
        { 
            for(i=s;i<=e;i++)
                {
                    for(j=s;j<=e;j++)
                        {
                            if(i==s||i==e||j==s||j==e)
                                {
                            a[i][j]=n;
                                }
                        }
                }
                         s++;
                         e--;
                         n--;
        }
        for(i=0;i<k;i++)
                {
                    for(j=0;j<k;j++)
                        {
                         printf("%d ",a[i][j]);
                        }
                        printf("\n");
                }
    return 0;
}

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