A reverse-order alphabet triangle prints a left-aligned staircase where each row’s highest letter grows forward, but letters along the row print backward down to A.
Remember
Rule: on row with peak i, print i down to A
A
BA
CBA
DCBA
EDCBA ← 5 rows
It is the row-order flip of Program 1: same growing widths, but the inner loop counts down instead of up. Compare with Program 3, which changes the start letter while still printing forward.
Approach
How to Solve It
Two ways to emit the same shape — start with nested char loops, then optionally reverse a substring.
Method
Idea
Best for
Nested char loops
Outer = peak letter; inner = peak down to A
Learning, interviews, exams
Reverse substring
Slice A..peak, reverse, print once
Shorter demos once loops click
Pseudocode
Pseudocode
for peak from 'A' to lastLetter:
for ch from peak down to 'A':
print ch (no newline)
print newline
Cheat sheet
Goal
Pattern
Walk each peak
for (char i = 'A'; i <= last; i++)
Print i..A
for (char j = i; j >= 'A'; j--) System.out.print(j);
Print letters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the reverse-order triangle updates instantly — capped at 26 letters (A–Z).
Whole numbers from 1 to 26. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 letters
A
BA
CBA
DCBA
EDCBA
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop peak letter and count how many times the inner loop prints.
Peak i
Inner j
Printed row
Letters
'A'
A..A
A
1
'B'
B..A
BA
2
'C'
C..A
CBA
3
'D'
D..A
DCBA
4
Total letter prints: 1 + 2 + 3 + 4 = 10 = 4×5/2. That triangular sum is why time is O(n²).
Code
Java Programs
Three complete programs: fixed end letter, Scanner input, and a reverse-substring shortcut. Use View Output to reveal sample results.
Example 1 — Fixed through 'E'
Hard-coded peak letter — outer grows A..E; inner counts down to A.
Java
public class ReverseOrderTriangle {
public static void main(String[] args) {
for (char i = 'A'; i <= 'E'; i++) {
for (char j = i; j >= 'A'; j--) {
System.out.print(j);
}
System.out.println();
}
}
}
Output
A
BA
CBA
DCBA
EDCBA
How It Works
1. Outer loop picks the peak.i runs from 'A' to 'E' — one row per peak letter.
2. Inner loop counts down. For each i, j runs from i down to 'A', so the row is i..A.
3. Print letters, then break the line.System.out.print(j) stays on the row; println() after the inner loop starts the next row.
When i = 'A' you get A; when i = 'B' you get BA; up through EDCBA.
Example 2 — User Input Version
Read the row count at runtime. Prefer validating with hasNextInt and clamping to 26 (shown in the tip below).
Java
import java.util.Scanner;
public class ReverseOrderTriangleInput {
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 last = (char)('A' + rows - 1);
for (char i = 'A'; i <= last; i++) {
for (char j = i; j >= 'A'; j--) {
System.out.print(j);
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
A
BA
CBA
DCBA
How It Works
1. Prompt and read. Ask for a row count, then read an int with Scanner.
2. Map rows to a last peak.last = (char)('A' + rows - 1) — for rows = 4, last is 'D'.
3. Same countdown core. Only the source of last changes — the print logic matches Example 1.
4. Safer input tip.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 — Reverse substring
Build A..peak with substring, reverse it, and print — same shape, no explicit inner countdown.
Java
public class ReverseOrderTriangleString {
public static void main(String[] args) {
int rows = 5;
String letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = 1; i <= rows; i++) {
StringBuilder sb = new StringBuilder(letters.substring(0, i));
System.out.println(sb.reverse());
}
}
}
Output
A
BA
CBA
DCBA
EDCBA
How It Works
1. Slice the forward prefix.substring(0, i) is A, AB, ABC, … — same as Program 1.
2. Reverse in place.StringBuilder.reverse() turns that prefix into A, BA, CBA, …
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.
j++
Forward triangle by mistake
If the inner loop increments from A to i, you reprint Program 1. Keep j-- from i down to 'A'.
Stop at B
Missing final A
Stopping before 'A' drops the tip letter every row. The condition must be j >= '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 — 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)
Reverse substring (Example 3)
O(rows²)
O(rows) per temporary StringBuilder
Total letters = 1 + 2 + … + n = n(n + 1)/2 — still quadratic in n. Same totals as Program 1; only print order along each row differs.
Remember
Key Takeaways
Rule: peak letter grows A..last; print that peak down to A.
Flip of Program 1: same widths — reverse the inner loop direction.
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 each peak i from 'A' to the last letter, print i down to 'A', then println.
Frequently Asked Questions
The outer loop selects the highest letter on the row (A to E). The inner loop starts at that letter and decrements down to A, printing each character, so each row reads backward.
Because the inner loop always runs until j equals A, so the last printed character on every row is A.
Program 1 prints forward along each row (A, AB, ABC). This pattern prints backward down to A (A, BA, CBA) while the left edge still grows A, B, C, …
Program 3 starts earlier each row but still prints forward to a fixed top (E, DE, CDE). This pattern starts later each row and prints backward to a fixed A.
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) for n rows, because the total printed letters are 1+2+…+n = n(n+1)/2.
Yes. Take letters.substring(0, i), reverse it with StringBuilder, and println the result. Nested char loops are better for learning; reverse is a handy shortcut later.
Check sc.hasNextInt() before sc.nextInt(), require n ≥ 1, and cap at 26 so bad input does not walk past Z.
🤔
Did you know?
Each row starts one letter later (A, B, C, …), but prints backward down to A. For 5 rows, the output is A, BA, CBA, DCBA, EDCBA — the reverse of Program 1’s row order.
