Printing Pattern Using Loops HackerRank Solution
Print a pattern of numbers from 1 to n as shown below. Each of the numbers is separated by a single space.
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 n.
Constraints
1<=n<=1000
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>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n;
scanf("%d", &n);
int len = 2*n - 1;
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
int min = i < j ? i : j;
min = min < len-i ? min : len-i-1;
min = min < len-j-1 ? min : len-j-1;
printf("%d ", n-min);
}
printf("\n");
}
return 0;
}
Steps used in solving the problem -
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Step 1: First we have imported required header files.
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Step 2: Then, we created the main function. we declared an integer variable n inside our function and used "scanf" function to read the user input.
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Step 3: Then, we defined the length and used two nested loops to iterate through i and j. this will print an square matrix with an odd number of rows and columns, based on the value of n.
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Step 4: At last, we printed the result.
