Reverse Triangle, E Always First in Java

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 Java 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 Java programs — fixed A–E, user-chosen row count, and a spaced-letter variant. Click View Output to reveal sample console results.

public class FixedStartReverseTriangle {
    public static void main(String[] args) {
        for (char i = 'A'; i <= 'E'; i++) {
            for (char j = 'E'; j >= i; j--) {
                System.out.print(j);
            }
            System.out.println();
        }
    }
}

EDCBA
EDCB
EDC
ED
E

import java.util.Scanner;

public class FixedStartReverseTriangleInput {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter the number of rows: ");
        int rows = sc.nextInt();

        char top = (char) ('A' + rows - 1);
        for (char i = 'A'; i <= top; i++) {
            for (char j = top; j >= i; j--) {
                System.out.print(j);
            }
            System.out.println();
        }

        sc.close();
    }
}

Enter the number of rows: 4
DCBA
DCB
DC
D

public class FixedStartReverseTriangleSpaced {
    public static void main(String[] args) {
        char top = 'E';

        for (char i = 'A'; i <= top; i++) {
            for (char j = top; j >= i; j--) {
                System.out.print(j + " ");
            }
            System.out.println();
        }
    }
}

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