Reverse Alphabet Decreasing Triangle in C

What You’ll Learn

Print a triangle where the first row is the full reverse run from 'E' down to 'A', and each next row becomes shorter: EDCBA, DCBA, CBA, BA, A. This is the reverse-direction companion to Program 6 (ABCDE, BCDE, … ending at E). Compare also with Program 8. Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count (or fixed top E), print a shrinking triangle where each row is a reverse run ending at A.

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

Inputs & Outputs

Minimal workflow

top = 'A' + rows - 1
for i from top down to 'A':     // start letter lowers
    for j from i down to 'A':   // reverse run each row
        print j
    print newline

Approach comparison

🔮 Live Preview

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

    return 0;
}

EDCBA
DCBA
CBA
BA
A

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

    return 0;
}

Enter the number of rows: 4
DCBA
CBA
BA
A

#include <stdio.h>

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

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

    return 0;
}

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

🔎 Worked Walkthrough — Top = E (5 rows)

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

Row lengths are 5, 4, 3, 2, 1. 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 n+(n-1)+…+1 = n(n+1)/2 characters, so total work is O(n²).

💡 Best Practices