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
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
C# Programs
Three complete programs: fixed end letter, console 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.
C#
using System;
class Program
{
static void Main()
{
for (char i = 'A'; i <= 'E'; i++)
{
for (char j = i; j >= 'A'; j--)
{
Console.Write(j);
}
Console.WriteLine();
}
}
}
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.Console.Write(j) stays on the row; WriteLine() 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 int.TryParse and clamp to 26 (shown in the tip below).
C#
using System;
class Program
{
static void Main()
{
Console.Write("Enter the number of rows: ");
int rows = Convert.ToInt32(Console.ReadLine());
char last = (char)('A' + rows - 1);
for (char i = 'A'; i <= last; i++)
{
for (char j = i; j >= 'A'; j--)
{
Console.Write(j);
}
Console.WriteLine();
}
}
}
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 convert the line to an int.
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.Convert.ToInt32 throws on letters. Prefer:
Safer input
if (!int.TryParse(Console.ReadLine(), out int rows) || rows < 1 || rows > 26)
{
Console.WriteLine("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.
C#
using System;
class Program
{
static void Main()
{
int rows = 5;
string letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = 1; i <= rows; i++)
{
char[] chars = letters.Substring(0, i).ToCharArray();
Array.Reverse(chars);
Console.WriteLine(new string(chars));
}
}
}
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.Array.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'.
WriteLine inside
Column of letters
If WriteLine is inside the inner loop, each letter lands on its own line. Use Write for letters; WriteLine 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
Use TryParse
Convert.ToInt32 throws on letters — prefer int.TryParse 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 row array
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 WriteLine 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 WriteLine.
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.
Console.Write stays on the same line. Console.WriteLine ends the current line. Letters use Write; the row break uses WriteLine 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 the characters, and WriteLine the result. Nested char loops are better for learning; reverse is a handy shortcut later.
Prefer int.TryParse(Console.ReadLine(), out rows) and clamp rows between 1 and 26 so bad input does not throw FormatException or 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.
