Inverted Incrementing Triangle Pattern

Understanding the Problem

The goal is to print an inverted triangle pattern where numbers increment sequentially, and each row has a decreasing number of elements.

Pattern Explanation

The pattern follows an inverted triangular structure where numbers increase sequentially from the bottom row to the top row.

7*8*9*10
4*5*6
2*3
1
            

Observations:

• The pattern consists of 4 rows.
• Numbers increment from left to right.
• Each row contains one less element than the previous row.
• Each number is followed by a '*', except the last number in each row.

Algorithm

The pattern follows these steps:

1. Initialize a number counter starting from 1.
2. Determine the total number of rows (N).
3. Calculate the starting number for the first row as (N * (N + 1)) / 2.
4. Loop through rows (from N down to 1).
5. Within each row, loop through columns (up to the current row number).
6. Print the number followed by '*' except for the last number in the row.
7. Increment the number after each print.
8. Move to the next line after printing each row.

Method 1: Using Nested Loops

This method uses two loops to print the pattern.

#include <stdio.h>

int main() {
    int n = 4;  // Number of rows
    int num = (n * (n + 1)) / 2;  // Starting number for the first row

    // Loop to print the inverted incrementing triangle pattern
    for (int i = n; i >= 1; i--) {  // Loop for each row
        for (int j = 1; j <= i; j++) {  // Loop for each column in the current row
            printf("%d", num++);  // Print the current number
            if (j < i) printf("*");  // Print '*' if not the last number in the row
        }
        printf("\n");  // Move to the next line after each row
    }

    return 0;
}
            

Output:

7*8*9*10
4*5*6
2*3
1