A reverse fixed-start alphabet triangle always begins each row at the same top letter and counts down, while the stopping letter rises so the reverse tail gets shorter.
Remember
Rule: for stop letter i from A to last,
print last down through i
EDCBA
EDCB
EDC
ED
E ← 5 rows (left edge fixed at E)
Same widths as Program 7 (5, 4, 3, 2, 1), but Program 7 moves the left edge (EDCBA, DCBA, …). Here the left edge stays put. Compare also with Program 5, which shrinks forward prefixes from A.
Approach
How to Solve It
Two ways to emit the same shape — start with nested letter loops, then optionally reverse a prefix once and take shorter leading slices.
Method
Idea
Best for
Nested letter loops
Outer = rising stop; inner = top..stop downward
Learning, interviews, exams
Reverse + top[:len]
Build EDCBA… once, take shorter prefixes
Shorter demos once loops click
Pseudocode
Pseudocode
top = lastLetter
for stop from 'A' to top:
for ch from top down to stop:
print ch (no newline)
print newline
Cheat sheet
Goal
Pattern
Raise stop letter
for i in range(ord('A'), top + 1):
Print top..i reverse
for j in range(top, i - 1, -1): print(chr(j), end="")
End the row
print()
Top letter from rows
top = ord('A') + rows - 1
One-line row shortcut
Reverse the A…top prefix, then top[:length] while length shrinks
Print letters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the reverse fixed-start 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
EDCBA
EDCB
EDC
ED
E
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop stop letter as i rises from 'A' to 'D' with top fixed at 'D'.
Stop i
Inner j
Printed row
Letters
'A'
D..A
DCBA
4
'B'
D..B
DCB
3
'C'
D..C
DC
2
'D'
D..D
D
1
Total letter prints: 4 + 3 + 2 + 1 = 10 = 4×5/2 — same triangular count as Programs 5 and 7.
Code
Python Programs
Three complete programs: fixed top letter, row-count input(), and a reverse-slice shortcut. Use View Output to reveal sample results.
Example 1 — Fixed top at 'E'
Hard-coded top letter — every row starts at E; the stop letter rises to shorten the tail.
Python
for i in range(ord('A'), ord('E') + 1):
for j in range(ord('E'), i - 1, -1):
print(chr(j), end="")
print()
Output
EDCBA
EDCB
EDC
ED
E
How It Works
1. Outer loop raises the stop.i runs from ord('A') to ord('E') — longest row first.
2. Inner loop always starts at E. For each stop, j runs from E down to i, so the row is E..i in reverse.
3. Print letters, then break the line.print(chr(j), end="") stays on the row; bare print() after the inner loop starts the next (shorter) row.
When i is A you get EDCBA; when i is E you get E.
Example 2 — User Input Version
Read the row count at runtime. Prefer try/except and clamp to 26 (shown in the tip below).
Python
rows = int(input("Enter the number of rows: "))
top = ord('A') + rows - 1
for i in range(ord('A'), top + 1):
for j in range(top, i - 1, -1):
print(chr(j), end="")
print()
Output (when user enters 4)
Enter the number of rows: 4
DCBA
DCB
DC
D
How It Works
1. Prompt and read. Ask for a row count, then convert the line to an int.
2. Map rows to a top letter.top = ord('A') + rows - 1 — for rows = 4, top is ord('D').
3. Same fixed-start core. Only the source of top changes — the print logic matches Example 1.
4. Safer input tip. Prefer:
Safer input
try:
rows = int(input("Enter the number of rows: "))
except ValueError:
print("Enter a whole number from 1 to 26.")
raise SystemExit
if rows < 1 or rows > 26:
print("Enter a whole number from 1 to 26.")
raise SystemExit
Example 3 — Reverse + top[:length]
Build the first reverse row once, then take shorter leading prefixes — same shape, no nested letter loop.
Python
rows = 5
letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
top = letters[:rows][::-1]
for length in range(rows, 0, -1):
print(top[:length])
Output
EDCBA
EDCB
EDC
ED
E
How It Works
1. Take and reverse the prefix.ABCDE reversed becomes EDCBA — the first printed row.
2. Shrink the leading slice.top[:5] is the full row; top[:4] drops the trailing A; and so on down to E.
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 = i
Program 7 by mistake
If the inner loop starts at i instead of top, you print EDCBA, DCBA, … Use for j in range(top, i - 1, -1).
stop at A
No shrinking
Stopping at A every time reprints the full reverse run. The stop must be the rising outer variable i.
print() early
Column of letters
If bare print() (or default print(chr(j))) sits inside the inner loop, each letter lands on its own line. Use end="" for letters; call print() only after the inner loop.
rows > 26
Past Z
ord('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 stop coincide. A good sanity check.
Bad input
ValueError on int()
Bare int(input()) raises on letters — prefer try/except ValueError and require 1–26.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(rows²)
O(1)
Reverse + slice (Example 3)
O(rows²)
O(rows) for the reversed prefix and temporary row strings
Total letters = n + (n - 1) + … + 1 = n(n + 1)/2 — quadratic in n. Same totals as Programs 5 and 7.
Remember
Key Takeaways
Rule: always start at the top letter; raise the stop i so each reverse row shortens.
vs Program 7: same widths — here the left edge stays fixed; there the left edge moves.
Break the row: call bare print() only after the inner loop.
Complexity:O(n²) time; O(1) extra space for nested loops.
One line: for stop i from A to the top letter, print top down through i, then print().
Frequently Asked Questions
Because the inner loop always starts at the top letter (E in the 5-row example) and counts down. So the first printed character each row is always E.
The outer loop increases the stopping point for the inner loop. That shortens the tail each row, producing EDCBA, then EDCB, then EDC, and so on.
Program 7 changes the first letter each row (E, then D, then C…). Program 8 keeps the first letter fixed and only shortens the reverse tail.
Program 5 prints forward prefixes from A (ABCDE, ABCD, …). This pattern prints reverse prefixes from a fixed top (EDCBA, EDCB, …). Same shrinking widths; opposite letter direction and left edge.
Every row would print the full reverse run (EDCBA each time) with no shrinking. The stop must be the rising outer variable.
print(chr(j), end="") stays on the same line. print() ends the current line. Letters use end=""; the row break uses print() after the inner loop.
O(n²) for n rows, because total printed characters are n(n+1)/2.
Yes. Reverse the A…top prefix, then print top[:length] while length shrinks from rows down to 1. Nested ord/chr loops are better for learning; slicing is a handy shortcut later.
Prefer try/except ValueError around int(input()) and clamp rows between 1 and 26 so bad input does not walk past Z.
🤔
Did you know?
Every row begins with the same top letter because the inner loop always starts there. The outer loop only raises the stopping point, so the tail shortens: EDCBA, EDCB, EDC, ED, E. Same widths as Program 7, but the left edge stays fixed.
