Incrementing + Inverted Squared Number-Star (Alternate)

Understanding the Problem

The goal is to print an alternate inverted squared number-star pattern where the row order alternates between normal and reverse.

Pattern Explanation

The pattern follows a grid structure where numbers increase sequentially, but alternate rows appear in reverse order.

1*2*3*4
9*10*11*12
5*6*7*8
13*14*15*16
            

Observations:

• The pattern consists of 4 rows and 4 columns.
• Numbers increment from left to right across the grid.
• Each number is followed by a '*', except the last number in each row.
• Even-indexed rows are printed normally, while odd-indexed rows are printed in reverse order.

Algorithm

The pattern follows these steps:

1. Initialize a number counter starting from 1.
2. Store values in a 2D array as they increment.
3. Loop through rows, printing normally for even rows and in reverse for odd rows.
4. Within each row, print numbers left to right with '*'.
5. Move to the next line after printing each row.

Method 1: Using Nested Loops

This method first stores values in an array and then prints them in alternating order.

#include <stdio.h>

void printPattern(int n) {
    int arr[n][n], num = 1;
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            arr[i][j] = num++;
        }
    }
    
    for (int i = 0; i < n; i++) {
        if (i % 2 == 0) {
            for (int j = 0; j < n; j++) {
                printf("%d", arr[i][j]);
                if (j < n - 1) printf("*");
            }
        } else {
            for (int j = n - 1; j >= 0; j--) {
                printf("%d", arr[i][j]);
                if (j > 0) printf("*");
            }
        }
        printf("\n");
    }
}

int main() {
    int size = 4;
    printPattern(size);
    return 0;
}
            

Output:

1*2*3*4
9*10*11*12
5*6*7*8
13*14*15*16