Print characters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the inverted forward triangle updates instantly — including letter and print counts.
Whole numbers from 1 to 26 (A–Z). Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · A–E · 15 letters
AAAAA
BBBB
CCC
DD
E
Trace
Worked Walkthrough — rows = 4 (A–D)
Trace each outer letter i and count how many times the inner loop prints it.
i
Inner j
Printed row
Count
A
D..A
AAAA
4
B
D..B
BBB
3
C
D..C
CC
2
D
D..D
D
1
Total letter prints: 4 + 3 + 2 + 1 = 10 = 4×5/2. That triangular sum is why time is O(n²).
Code
Java Programs
Three complete programs: fixed A–E, Scanner input, and a String.repeat shortcut. Use View Output to reveal sample results.
Example 1 — Fixed 'A' up to 'E'
Hard-coded range — ideal for first demos and screenshots.
Java
public class InvertedForward {
public static void main(String[] args) {
for (char i = 'A'; i <= 'E'; i++) {
for (char j = 'E'; j >= i; j--) {
System.out.print(i);
}
System.out.println();
}
}
}
Output
AAAAA
BBBB
CCC
DD
E
How It Works
1. Outer loop picks the letter.i runs from 'A' to 'E' — that letter fills the whole row.
2. Inner loop sets the width.j runs from 'E' down to i, so widths are 5, 4, 3, 2, 1.
3. Print i, not j.System.out.print(i) keeps the row uniform (AAAAA, not EDCBA).
4. Break the line.System.out.println() after the inner loop starts the next row.
Example 2 — User Input Version
Read the height at runtime. Prefer hasNextInt() and clamp to 1–26 (shown in the tip below).
Java
import java.util.Scanner;
public class InvertedForwardInput {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of rows: ");
int rows = sc.nextInt();
for (int row = 0; row < rows; row++) {
char ch = (char)('A' + row);
int repeat = rows - row;
for (int k = 1; k <= repeat; k++) {
System.out.print(ch);
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
AAAA
BBB
CC
D
How It Works
1. Prompt and read. Ask for a row count, then store it with sc.nextInt().
2. Index form of the same shape. Row 0 prints Arows times; row 1 prints Brows - 1 times; and so on.
3. Safer input tip. Unchecked nextInt() throws on letters. Prefer:
Safer input
if (!sc.hasNextInt()) {
System.out.println("Enter a whole number from 1 to 26.");
return;
}
int rows = sc.nextInt();
if (rows < 1 || rows > 26) {
System.out.println("Enter a whole number from 1 to 26.");
return;
}
Example 3 — String.valueOf(ch).repeat(repeat)
Build each repeated-letter row in one call — same shape, no explicit inner print loop (Java 11+).
Java
public class InvertedForwardString {
public static void main(String[] args) {
int rows = 5;
for (int row = 0; row < rows; row++) {
char ch = (char)('A' + row);
int repeat = rows - row;
System.out.println(String.valueOf(ch).repeat(repeat));
}
}
}
Output
AAAAA
BBBB
CCC
DD
E
How It Works
1. One outer loop. Still walk row from 0 to rows - 1.
2. Build the row.String.valueOf(ch).repeat(repeat) creates a string of length repeat filled with ch.
3. Print and advance.println prints that string and ends the line.
Learn the two-loop version first (Examples 1–2) so you can explain both bounds in an interview; treat this as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
print(j)
Stepping letters
If you print j instead of i, the first row becomes EDCBA. Always System.out.print(i) for this shape.
Countdown outer
Program 11 by mistake
Counting i from top down to A prints EEEEE first. Outer loop must advance A→top.
println inside
Column of letters
If println sits inside the inner loop, each letter lands on its own line. Call it only after the inner loop.
rows > 26
Past Z
Clamp to 1–26 so 'A' + row never walks past Z.
rows = 1
Single A
Output is just A — a good sanity check.
Bad input
Check hasNextInt
nextInt() throws on letters — prefer hasNextInt() and require 1–26.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(rows²)
O(1)
String.repeat (Example 3)
O(rows²)
O(rows) per temporary row string
Total letters printed = n + (n-1) + … + 1 = n(n+1)/2, which is still quadratic in n.
Remember
Key Takeaways
Rule: letters advance A→top; widths shrink n..1.
Print i: the outer letter fills the row; j only sets the count.
Break the row:print for letters; println after the inner loop.
Complexity:O(n²) time from the triangular letter count; O(1) extra space for nested loops.
One line: for each letter i from A to top, print i(top - i + 1) times, then println.
Frequently Asked Questions
Program 11 prints EEEEE, DDDD, ... (letters step down). Program 12 prints AAAAA, BBBB, ... (letters step up) with the same inverted widths 5..1.
When i is A, the inner loop runs from E down to A (5 times) and prints A each time. Next i becomes B, the inner loop runs 4 times (E..B) and prints B.
j only controls how many times the loop runs. Printing i keeps the entire row the same letter; printing j would step letters across the row.
Because the inner loop runs from the fixed top letter down to the current i. As i increases, the loop has fewer iterations.
System.out.print stays on the same line. System.out.println ends the current line. Letters use print; the row break uses println after the inner loop.
O(n²) where n is the number of rows. Total System.out.print calls equal n+(n-1)+…+1 = n(n+1)/2.
Yes. System.out.println(String.valueOf(ch).repeat(repeat)) prints a full repeated-letter row in one call (Java 11+). Nested loops are better for learning; String.repeat is a handy shortcut later.
Check sc.hasNextInt() before sc.nextInt() and clamp rows between 1 and 26 so bad input does not throw InputMismatchException or walk past Z.
🤔
Did you know?
This is the forward-letter twin of Program 11: same inverted widths (5…1), but letters advance A→E instead of stepping down. Print the outer loop letter inside the inner loop so each row stays uniform.
