Reverse Triangle, E Always First in C

What You’ll Learn

Print a reverse triangle where every row begins with the same top letter (E in the 5-row example) and becomes shorter each line: EDCBA, EDCB, EDC, ED, E. Compare Program 7 (the first letter changes each row) and Program 5 (forward prefixes from A). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count (or fixed top E), print a shrinking reverse triangle where every row begins at the same top letter.

// Five rows (top = E)
// EDCBA
// EDCB
// EDC
// ED
// E

Inputs & Outputs

Minimal workflow

top = 'A' + rows - 1
for i from 'A' to top:          // raise the floor
    for j from top down to i:   // reverse from fixed start
        print j
    print newline

Approach comparison

🔮 Live Preview

Choose 1–10 rows and draw the fixed-start reverse 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 = 'E'; j >= i; j--) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

EDCBA
EDCB
EDC
ED
E

#include <stdio.h>

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

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

    top = (char)('A' + rows - 1);
    for (i = 'A'; i <= top; i++) {
        for (j = top; j >= i; j--) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

Enter the number of rows: 4
DCBA
DCB
DC
D

#include <stdio.h>

int main() {
    char top = 'E';
    char i, j;

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

    return 0;
}

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

🔎 Worked Walkthrough — Top = E (5 rows)

Trace each floor letter and the resulting reverse prefix from E.

Row lengths are 5, 4, 3, 2, 1. The left edge is always E.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

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

💡 Best Practices