An inverted repeating alphabet triangle prints a wide first row of one letter, then shrinks by one repeated letter on each new line while letters count down.
Remember
Rule: for letter i from top down to 'A',
print i exactly (i - 'A' + 1) times
EEEEE
DDDD
CCC
BB
A ← 5 rows (top = 'E')
It is the flip of Program 10: same countdown letters, but widths shrink 5, 4, 3, …, 1 instead of growing. Print the outer letter inside the inner loop so each row stays uniform.
Approach
How to Solve It
Two ways to emit the same shape — start with nested char loops, then optionally shorten with String.repeat.
Method
Idea
Best for
Nested char loops
Outer = letter; inner = width; print outer letter
Learning, interviews, exams
String.valueOf(ch).repeat(n)
Build a whole repeated-letter row in one call
Shorter demos once loops click (Java 11+)
Pseudocode
Pseudocode
for i from top down to 'A':
for j from 'A' to i:
print i (no newline)
print newline
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the height and the inverted triangle updates instantly — including the letter total.
Whole numbers from 1 to 26 (A–Z). Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 letters
EEEEE
DDDD
CCC
BB
A
Trace
Worked Walkthrough — 'E' down to 'A'
Trace each outer-loop value of i and count how many times the inner loop runs.
i
Inner j range
Printed row
Repeats
'E'
'A'..'E'
EEEEE
5
'D'
'A'..'D'
DDDD
4
'C'
'A'..'C'
CCC
3
'B'
'A'..'B'
BB
2
'A'
'A'..'A'
A
1
Total letter prints: 5 + 4 + 3 + 2 + 1 = 15 = 5×6/2. That triangular sum is why time is O(n²).
Code
Java Programs
Three complete programs: fixed letters, Scanner input, and a String.repeat shortcut. Use View Output to reveal sample results.
Example 1 — Fixed 'E' down to 'A'
Hard-coded range — outer letter counts down; inner loop shrinks and prints that letter.
Java
public class InvertedRepeat {
public static void main(String[] args) {
for (char i = 'E'; i >= 'A'; i--) {
for (char j = 'A'; j <= i; j++) {
System.out.print(i);
}
System.out.println();
}
}
}
Output
EEEEE
DDDD
CCC
BB
A
How It Works
1. Outer loop picks the letter.i runs from 'E' down to 'A'.
2. Inner loop sets the width.j runs from 'A' to i — five times for E, four for D, and so on.
3. Print the outer letter.System.out.print(i) keeps the whole row the same character.
4. Break the line.System.out.println() after the inner loop starts the next row.
Example 2 — User Input Version
Compute top = 'A' + rows - 1, then shrink with repeat = ch - 'A' + 1. Prefer hasNextInt() in real apps.
Java
import java.util.Scanner;
public class InvertedRepeatInput {
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 ch = top; ch >= 'A'; ch--) {
int repeat = ch - 'A' + 1;
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
DDDD
CCC
BB
A
How It Works
1. Prompt and read. Ask for a row count, then store it with sc.nextInt().
2. Compute the top letter. For rows = 4, top is 'D' — first row is DDDD.
3. Same shrink core.repeat = ch - 'A' + 1 gives widths 4, 3, 2, 1 as ch counts down.
4. 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 character loop (Java 11+).
Java
public class InvertedRepeatString {
public static void main(String[] args) {
int rows = 5;
char top = (char)('A' + rows - 1);
for (char ch = top; ch >= 'A'; ch--) {
int repeat = ch - 'A' + 1;
System.out.println(String.valueOf(ch).repeat(repeat));
}
}
}
Output
EEEEE
DDDD
CCC
BB
A
How It Works
1. Same countdown. Still walk ch from top down to 'A'.
2. Build the row.String.valueOf(ch).repeat(repeat) creates a string of length repeat filled with ch.
3. Learn loops first. Use Examples 1–2 when you need to show nested bounds; treat this as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
print(j)
Stepping letters on a row
If you print j instead of i, you get ABCDE on the first row. Print the outer letter to keep each row uniform.
i++
Growing triangle by mistake
Counting letters upward with growing widths reprints Program 10’s shape family. Keep i-- from the top letter down to 'A'.
println inside
Column of letters
If println is inside the inner loop, each letter lands on its own line. Use print for letters; println only after the inner loop.
rows > 26
Past Z
'A' + rows - 1 leaves A–Z when rows > 26. Clamp or reject in interactive programs.
rows = 1
Single A
Output is just A — top and tip coincide. 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 = n + (n - 1) + … + 1 = n(n + 1)/2 — still quadratic in n. Same totals as Program 10; only the print order of widths differs.
Remember
Key Takeaways
Rule: countdown letter i; print i exactly i - 'A' + 1 times.
Flip of Program 10: same letters — widths shrink instead of grow.
Break the row: call println only after the inner loop.
Complexity:O(n²) time from the triangular letter count; O(1) extra space for nested loops.
One line: for i from the top letter down to 'A', print i for every j from 'A' to i, then println.
Frequently Asked Questions
Program 10 prints widths 1..5 (E, DD, CCC, ...). Program 11 prints widths 5..1 (EEEEE, DDDD, ...). Both use repeating letters per row and letters counting down.
When i is E, the inner loop runs from A through E (5 times) and prints E each time. Next i becomes D, the inner loop runs 4 times (A..D) and prints D.
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 stops at j <= i. Smaller i means fewer iterations, so fewer characters are printed.
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 inverted twin of Program 10: letters still step E→A, but widths shrink 5, 4, 3, …, 1 instead of growing. Print the outer loop letter inside the inner loop so each row stays uniform.
