Printing Pattern using Loops - Hackerrank Solution

Printing Pattern using Loops - Hackerrank Solution

Objective

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 

SOLUTION

#include <stdio.h> main() { int n, l, i, j, a, b, c, d, minimum, x; scanf("%d", &n); l = (2 * n) - 1; for (i = 0; i < l; i++) { for (j = 0; j <= l - 1; j++) { a = j; b = i; c = l - 1 - j; d = l - 1 - i; if (a < b) x = a; else x = b; if (c < x) x = c; if (d < x) x = d; printf("%d ", n - x); } printf("\n"); } }

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