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  1. Prepare
  2. C
  3. Conditionals and Loops
  4. Printing Pattern Using Loops
  5. Discussions

Printing Pattern Using Loops

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1043 Discussions

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  • macohip968
     + 0 comments

    I remember trying to solve a similar pattern problem when I was first learning programming. It reminded me of الهندسة المعمارية, where every layer builds upon the previous one to create a structured and symmetrical design. The logic behind decreasing and increasing numbers in a pattern felt just like designing a blueprint—precision and symmetry matter the most!

  • galant4456
     + 0 comments

    include

    include

    include

    include

    int min(int i, int j, int k, int x){ int minimum = i;

    if(j < minimum){
    minimum = j;
} 
if(k < minimum){
    minimum = k;
} 
if(x < minimum){
    minimum = x;
} 

return minimum;

    }

    int main() {

    int n, distance, value;
scanf("%d", &n);
// Complete the code to print the pattern.
int size = 2 * n - 1;
for(int i = 0; i < size; i++){
    for(int j = 0; j < size ; j++){
        distance = min(i, j, size - i - 1, size - j - 1);
        value = n - distance;
        printf("%d ", value);
    }
    printf("\n");
}

return 0;

    }

  • girigh1998
     + 0 comments
    void printPattern(int n) {
    int size = 2 * n - 1;
    for (int i = 0; i < size; i++) {
        for (int j = 0; j < size; j++) {
            int min = i < j ? i : j;
            min = min < size - i ? min : size - i - 1;
            min = min < size - j ? min : size - j - 1;
            printf("%d ", n - min);
        }
        printf("\n");
    }
}

int main() {
    int n;
    scanf("%d", &n);
    printPattern(n);
    return 0;
}

  • zero006607578
     + 0 comments

    Simple Understandable Approach

    int size = 2 * n - 1;

for (int i = 0; i < size; i++) {
    for (int j = 0; j < size; j++) {
        for (int k = 0; k < n; k++) {
            if (i == k || i == size - k - 1 || j == k || j == size - k - 1) {
                printf("%d ", n - k);
                break;
            }
        }
    }
    printf("\n");
}

    just going layer by layer

    if(i == k || ...

    print...

    break; // this is the most important part cause it only allows one layer at a time

  • tharoonsays
     + 0 comments
    #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n,k;
    scanf("%d",&n);
    for(int i=0; i<n; i++){
        k = n;
        for(int j=0; j<2*n-1; j++) {
            printf("%d ",k);
            k = (((k-j)>(k-i)) && (j < 2*(n-1)-i)) ? k-1 : k;
            k = ((k < n) && (j >= 2*(n-1)-i)) ? k+1 : k;
        }
        printf("\n");
    }
    for(int i=n-2; i>-1; i--){
        k = n;
        for(int j=0; j<2*n-1; j++) {
            printf("%d ",k);
            k = (((k-j)>(k-i)) && (j < 2*(n-1)-i)) ? k-1 : k;
            k = ((k < n) && (j >= 2*(n-1)-i)) ? k+1 : k;
        }
				
				This code mimics signal processing and stuff used in Digital Systems, Ive used concept of DEMUX and split the loop conditions 

        printf("\n");
}
return 0;

    }