Reverse Order Alphabet Triangle in C

What You’ll Learn

Print a right-angled triangle where the first character of each row moves forward (A, B, C, D, E), but each row prints letters in reverse order down to A: A, BA, CBA, DCBA, EDCBA. Compare with Program 1 (forward along each row) and Program 3 (reverse starting letter, forward along row). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count (or fixed top E), print a right-angled triangle where each row starts at a higher letter and counts down to A.

// Five rows (top = E)
// A
// BA
// CBA
// DCBA
// EDCBA

Inputs & Outputs

Minimal workflow

top = 'A' + rows - 1
for i from 'A' to top:          // peak letter grows
    for j from i down to 'A':   // reverse along the row
        print j
    print newline

Approach comparison

🔮 Live Preview

Choose 1–10 rows and draw the reverse-order alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed A–E, scanf row count, and a spaced-letter variant. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    char i, j;

    for (i = 'A'; i <= 'E'; i++) {
        for (j = i; j >= 'A'; j--) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

A
BA
CBA
DCBA
EDCBA

#include <stdio.h>

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

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

    for (i = 'A'; i < 'A' + rows; i++) {
        for (j = i; j >= 'A'; j--) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

Enter the number of rows: 4
A
BA
CBA
DCBA

#include <stdio.h>

int main() {
    char i, j;

    for (i = 'A'; i <= 'E'; i++) {
        for (j = i; j >= 'A'; j--) {
            printf("%c ", j);
        }
        printf("\n");
    }

    return 0;
}

A 
B A 
C B A 
D C B A 
E D C B A 

🔎 Worked Walkthrough — Top = E (5 rows)

Trace each peak letter and the resulting reverse run down to A.

Row lengths are 1, 2, 3, 4, 5. The right edge is always A.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For n rows you print 1+2+…+n = n(n+1)/2 characters, so total work is O(n²).

💡 Best Practices