Descending Numbers with Diagonal Asterisk in C

What You’ll Learn

The descending number pattern with diagonal asterisk prints 5432*, 543*1, 54*21, 5*321, *4321 — a natural step after the bidirectional triangle in Program 25. This tutorial covers descending digits, the i == j condition, a live preview, algorithm steps, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer n (e.g. 5), print n rows of descending digits with one diagonal * per row where i == j.

// n = 5 (conceptual shape)
// 5432*
// 543*1
// 54*21
// 5*321
// *4321

Inputs & Outputs

Minimal workflow

for i from 1 to n:
    for j from n down to 1:
        if i == j:
            print "*"
        else:
            print j
    print newline

Approach comparison

🔮 Live Preview

Choose a size between 3 and 9 and draw the descending diagonal asterisk pattern in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed size, custom symbol, and user input. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    int i, j;

    for (i = 1; i <= 5; ++i) {
        for (j = 5; j >= 1; --j) {
            if (i == j)
                printf("*");
            else
                printf("%d", j);
        }
        printf("\n");
    }

    return 0;
}

5432*
543*1
54*21
5*321
*4321

#include <stdio.h>

int main() {
    int n = 5;
    int i, j;

    for (i = 1; i <= n; ++i) {
        for (j = n; j >= 1; --j) {
            if (i == j)
                printf("#");
            else
                printf("%d", j);
        }
        printf("\n");
    }

    return 0;
}

5432#
543#1
54#21
5#321
#4321

#include <stdio.h>

int main() {
    int n;
    int i, j;

    printf("Enter size: ");
    scanf("%d", &n);

    for (i = 1; i <= n; ++i) {
        for (j = n; j >= 1; --j) {
            if (i == j)
                printf("*");
            else
                printf("%d", j);
        }
        printf("\n");
    }

    return 0;
}

Enter size: 4
432*
43*1
4*21
*321

🔎 Worked Walkthrough — n = 5

Trace each outer-loop value of i, where i == j, and the full row output.

The star position moves left as i increases — each row still prints exactly n characters.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices