Bidirectional Number Triangle in C

What You’ll Learn

The bidirectional number triangle prints 11111, 2222, 333, 22, 1 — a natural step after the centered pyramid in Program 24. This tutorial covers shrinking rows, if/else digit mapping, a live preview, algorithm steps, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a positive integer rows (e.g. 5), print a shrinking triangle where each row repeats one digit — rising on early rows, mirroring down on the last rows.

// rows = 5 (conceptual shape)
// 11111
// 2222
// 333
// 22
// 1

Inputs & Outputs

Minimal workflow

for i from 1 to rows:
    if i < rows - 1:
        val = i
    else:
        val = rows + 1 - i
    for j from i to rows:
        print val
    print newline

Approach comparison

🔮 Live Preview

Choose a row count between 3 and 9 and draw the bidirectional number triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed rows, user input, and spaced output variant. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    int i, j;

    for (i = 1; i <= 5; ++i) {
        for (j = i; j <= 5; ++j) {
            if (i < 4)
                printf("%d", i);
            else
                printf("%d", 6 - i);
        }
        printf("\n");
    }

    return 0;
}

11111
2222
333
22
1

#include <stdio.h>

int main() {
    int rows;
    int i, j, val;

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

    for (i = 1; i <= rows; ++i) {
        val = (i < rows - 1) ? i : (rows + 1 - i);

        for (j = i; j <= rows; ++j)
            printf("%d", val);

        printf("\n");
    }

    return 0;
}

Enter rows: 5
11111
2222
333
22
1

#include <stdio.h>

int main() {
    int rows = 5;
    int i, j, val;

    for (i = 1; i <= rows; ++i) {
        val = (i < rows - 1) ? i : (rows + 1 - i);

        for (j = i; j <= rows; ++j)
            printf("%d ", val);

        printf("\n");
    }

    return 0;
}

1 1 1 1 1 
2 2 2 2 
3 3 3 
2 2 
1 

🔎 Worked Walkthrough — rows = 5

Trace each outer-loop value of i, the digit chosen, inner-loop count, and row output.

Prints per row = rows - i + 1 — total prints = n(n+1)/2 for n rows.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices