A reverse-starting-letter alphabet triangle grows leftward: each row begins one letter earlier, but letters along the row still run forward to a fixed top.
Remember
Rule: start letter moves top→A; print forward to top
E
DE
CDE
BCDE
ABCDE ← top = 'E' (5 rows)
Mixes Program 1 (forward run) with Program 2 (moving start). The right edge stays fixed at top on every row.
Approach
How to Solve It
Two ways to emit the same shape — nested char loops, or a growing suffix of a fixed alphabet string.
Method
Idea
Best for
Nested char loops
Outer = start letter; inner = start up to top
Learning, interviews, exams
Alphabet suffix
Slice the last r letters of "ABC…top"
Shorter demos once loops click
Pseudocode
Pseudocode
top = 'A' + rows - 1
for start from top down to 'A':
for ch from start up to top:
print ch (no newline)
print newline
Change the row count and the reverse-start forward triangle updates instantly — every row ends at the same top letter.
Whole number from 1 to 10. Tap a chip or type a count — the preview redraws as you go.
Live result5 rows · top E · 15 letters
E
DE
CDE
BCDE
ABCDE
Trace
Worked Walkthrough — Top = E (5 rows)
Trace each start letter and the resulting forward suffix.
i (start)
Inner range
Printed row
E
E..E
E
D
D..E
DE
C
C..E
CDE
B
B..E
BCDE
A
A..E
ABCDE
Row lengths are 1, 2, 3, 4, 5. The right edge is always E.
Code
C# Programs
Three complete programs: fixed top E, row-count input, and spaced letters. Use View Output to reveal sample results.
Example 1 — Fixed Top E
Outer loop chooses the first letter on the row; inner loop prints forward up to 'E'.
C#
using System;
class Program
{
static void Main()
{
for (char i = 'E'; i >= 'A'; i--)
{
for (char j = i; j <= 'E'; j++)
{
Console.Write(j);
}
Console.WriteLine();
}
}
}
Output
E
DE
CDE
BCDE
ABCDE
How It Works
1. Outer loop moves the start.i runs from 'E' down to 'A' — one start letter per row.
2. Inner loop prints forward. For each row, j runs from i up to 'E'. When i = 'C', that prints CDE.
3. Break the line.WriteLine() after the inner loop ends the row.
4. Full last row. When i = 'A', the inner loop prints the full forward run ABCDE.
Example 2 — Row Count Input
Read the number of rows and compute top = 'A' + rows - 1. Prefer int.TryParse in real apps.
C#
using System;
class Program
{
static void Main()
{
Console.Write("Enter the number of rows: ");
int rows = Convert.ToInt32(Console.ReadLine());
char top = (char)('A' + rows - 1);
for (char i = top; i >= 'A'; i--)
{
for (char j = i; j <= top; j++)
{
Console.Write(j);
}
Console.WriteLine();
}
}
}
Output (when user enters 4)
Enter the number of rows: 4
D
CD
BCD
ABCD
How It Works
1. Scale the top. For 4 rows, top becomes 'D'. Cap rows at 26 so top stays within A–Z.
2. Same core loops. Only the stop letter changes; the start-moves-backward rule is unchanged.
3. Safer input tip. 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 — Spaced Letters
Print a trailing space after each letter so columns are easier to scan.
C#
using System;
class Program
{
static void Main()
{
char top = 'E';
for (char i = top; i >= 'A'; i--)
{
for (char j = i; j <= top; j++)
{
Console.Write(j + " ");
}
Console.WriteLine();
}
}
}
Output
E
D E
C D E
B C D E
A B C D E
How It Works
1. Same bounds. Loop limits are unchanged — only the printed unit becomes j + " ".
2. Optional trim. Trim trailing spaces later if you need a compact line for grading scripts.
Edge Cases & Pitfalls
Check these before calling the solution done.
Inner j--
Wrong direction on the row
If the inner loop counts down, you get Program 2 (E, ED, EDC). Keep j++ from i to top.
Outer ++
Growing from A instead
If the outer loop runs A..top with the same inner rule, you print Program 1 upside-down. Start at top and go down to 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 row.
rows = 1
Single A
When top = 'A', you print one row: A. A good sanity check.
rows > 26
Beyond Z
Cap at 26 (or reject) so top stays within A–Z.
Bad input
Use TryParse
Convert.ToInt32 throws on empty or non-numeric text. Prefer int.TryParse and a clear message.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(n²)
O(1)
Spaced letters (Example 3)
O(n²)
O(1)
For n rows, total letters are 1 + 2 + … + n = n(n + 1)/2 — still quadratic in n.
Remember
Key Takeaways
Rule: start letter moves top → A; print forward to top.
Fixed right edge: every row ends at the same top letter.
Break the row: call WriteLine only after the inner loop.
Complexity:O(n²) time; O(1) extra space.
One line: for each start from top down to A, print that letter forward up to top.
Frequently Asked Questions
The outer loop moves the starting letter from E down to A. The inner loop always prints forward from that start letter up to E (or top), so the last character remains E on every row.
Program 2 prints letters descending along the row (E, ED, EDC). This program prints forward along the row (E, DE, CDE) while only the row’s first letter moves backward.
Program 1 always starts at A and grows the end letter (A, AB, ABC). This pattern grows the start letter backward while keeping the right edge fixed at top.
Yes. Read rows from the console and set top = (char)('A' + rows - 1). Then loop i from top down to 'A' and print j from i up to top.
O(n²) for n rows, because the total printed letters are 1+2+...+n = n(n+1)/2.
Prefer int.TryParse, require n ≥ 1, and cap at 26 so the top letter stays within A–Z.
Because the inner loop always stops at top (E in the fixed example), so the last printed character is always that same letter.
Yes. Use 'a' as the base: top = (char)('a' + rows - 1), then loop the same way with j++ up to top.
🤔
Did you know?
This triangle changes only the starting letter of each row (E, D, C, …), while letters along the row still increase forward. In the 5-row example, every row ends at E, producing E, DE, CDE, BCDE, ABCDE.
