Python Palindromic Alphabet Pyramid Pattern (Center A)
Beginner
7 min read
Updated: Sep 2026
3 programs
Live preview
Definition
What Is This Pattern?
A palindromic alphabet pyramid prints mirrored letter rows that grow around a single center A: each line reads the same forward and backward.
Remember
Rule: for peak i from 'A' to top,
print i..B (descending), then A..i (ascending)
A
BAB
CBABC
DCBABCD
EDCBABCDE ← 5 rows (top = 'E')
The descending wing stops before A so the center letter is not printed twice. Compare with Program 18 (different wing order) and add pads (or see Program 16) if you want the shape centered.
Approach
How to Solve It
Two ways to emit the same palindrome — classic ord/chr loops, then optionally space each letter for a wider look.
Method
Idea
Best for
Descend + ascend
Left wing i..B; right wing A..i
Learning, interviews, exams
Spaced letters
Same wings; print(chr(j), end=" ")
Clearer demos in dense terminals
Pseudocode
Pseudocode
for i from 'A' to top:
for j from i down to 'B':
print j (no newline)
for j from 'A' to i:
print j (no newline)
print newline
Cheat sheet
Goal
Pattern
Pick the peak
for i in range(ord("A"), ord("E") + 1):
Left wing
for j in range(i, ord("A"), -1): print(chr(j), end="")
Center + right
for j in range(ord("A"), i + 1): print(chr(j), end="")
End the row
print()
Row length
2 * (i - ord("A")) + 1 (odd widths)
Peak from rows
peak = ord("A") + r
Avoid double A
Use range(i, ord("A"), -1) — stops before A
Printing Letters vs Starting a New Line
API
Effect
Use for
print(..., end="")
Stays on the same line
Each letter on both wings
print()
Ends the current line
After both inner loops
Print letters without a newline, then end the row once.
Try it
Live Preview
Change the height and the palindrome updates instantly — including the letter total (n²).
Whole numbers from 1 to 10. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 25 letters
A
BAB
CBABC
DCBABCD
EDCBABCDE
Trace
Worked Walkthrough — A–E
Trace each row’s left wing, center + right wing, and full line.
i
Left (i..B)
Right (A..i)
Printed row
A
(empty)
A
A
B
B
AB
BAB
C
CB
ABC
CBABC
D
DCB
ABCD
DCBABCD
E
EDCB
ABCDE
EDCBABCDE
Lengths: 1 + 3 + 5 + 7 + 9 = 25 = 5². That odd-number sum is why time is O(n²).
Code
Python Programs
Three complete programs: fixed A–E, input() row-count, and a spaced-letter variant. Use View Output to reveal sample results.
Example 1 — Fixed A–E
Two loops per row: descending (i..B), then ascending (A..i). The descending loop stops before A.
Python
for i in range(ord("A"), ord("E") + 1):
for j in range(i, ord("A"), -1):
print(chr(j), end="")
for j in range(ord("A"), i + 1):
print(chr(j), end="")
print()
Output
A
BAB
CBABC
DCBABCD
EDCBABCDE
How It Works
1. Outer loop picks the peak.i runs from ord("A") to ord("E") — one row per peak letter.
2. Left wing descends.range(i, ord("A"), -1) prints down to B and stops before A.
3. Center + right ascend. Print from A through i — that supplies the single center A.
4. Break the line. Bare print() after both wings starts the next row.
When i is C, the left wing prints CB and the right wing prints ABC → CBABC.
Example 2 — Row Count Input
Cap at 26 for A–Z. Prefer try/except ValueError in real apps.
Python
n = int(input("Enter number of rows (1..26): "))
if n < 1:
raise SystemExit
if n > 26:
n = 26
for r in range(n):
peak = ord("A") + r
for ch in range(peak, ord("A"), -1):
print(chr(ch), end="")
for ch in range(ord("A"), peak + 1):
print(chr(ch), end="")
print()
Output (when user enters 4)
Enter number of rows (1..26): 4
A
BAB
CBABC
DCBABCD
How It Works
1. Prompt and clamp. Read a row count and keep it in 1–26 so peaks stay in A–Z.
2. Map index to peak. Row r uses peak = ord("A") + r.
3. Same wing core. Descend/ascend rules match Example 1 — only the number of peaks changes.
4. Safer input tip. Prefer:
Safer input
try:
n = int(input("Enter number of rows (1..26): "))
except ValueError:
print("Enter a whole number from 1 to 26.")
raise SystemExit
if n < 1 or n > 26:
print("Enter a whole number from 1 to 26.")
raise SystemExit
Example 3 — Spaced Letters
Same palindrome with a space after each letter for a wider look.
Python
for i in range(ord("A"), ord("E") + 1):
for j in range(i, ord("A"), -1):
print(chr(j), end=" ")
for j in range(ord("A"), i + 1):
print(chr(j), end=" ")
print()
Output
A
B A B
C B A B C
D C B A B C D
E D C B A B C D E
How It Works
1. Same wings. Still descend i..B, then ascend A..i.
2. Format only.end=" " adds a trailing space after each letter.
3. Optional polish. Trim the final space per row if you need a clean line end.
Edge Cases & Pitfalls
Check these before calling the solution done.
Through A
Duplicate center A
If the descending loop goes through A (e.g. range(i, ord("A") - 1, -1)), you get BAAB or CBABBC. Keep range(i, ord("A"), -1).
Wing order
Wrong palindrome style
Ascending first then descending gives Program 18-style rows (e.g. ABCBA) instead of CBABC.
print() inside
Column of letters
If bare print() (or default newline) is inside either wing loop, each letter lands on its own line. Use end="" for letters; call print() only after both loops.
n = 1
Single A
Left wing never runs; output is just A. A good sanity check.
n > 26
Past Z
Clamp or reject so peaks stay in A–Z.
Bad input
Use try/except
int(input()) raises on letters — prefer try/except ValueError and require 1–26.
Analysis
Time and Space Complexity
Program
Time
Extra space
Descend + ascend (Examples 1–2)
O(n²)
O(1)
Spaced letters (Example 3)
O(n²)
O(1)
Total letters = 1 + 3 + … + (2n - 1) = n². Spaced output prints the same letters plus spaces — still quadratic in n.
Remember
Key Takeaways
Rule: for peak i, print i..B then A..i.
Single center A: use range(i, ord("A"), -1) so the descending wing stops before A.
Break the row: call print() only after both wings.
Complexity:O(n²) time from n² letters; O(1) extra space.
One line: for each peak i, print descending i..B, then ascending A..i, then print().
Frequently Asked Questions
The first loop prints descending letters from the row peak down to B. The second prints ascending from A through i. Together they mirror around a single A.
range(i, ord('A'), -1) stops before A. The forward loop already prints A, so stopping the descending loop before A avoids duplicating the center character.
The descending loop stops before A, and the ascending loop starts at A. That prevents printing A twice at the join.
For row letter i, length is 2*(i-ord('A'))+1, so rows grow as 1, 3, 5, 7, 9 for A..E.
Change the outer loop upper bound from ord('E') to your target letter (like ord('H')), or use the row-count input() variant.
Program 18 typically builds A..peak then peak-1..A. This pattern descends to B first, then ascends A..i, still mirroring around a single A.
O(n²) for n rows because total printed characters are 1+3+...+(2n-1) = n².
No — the classic version is left-aligned with no leading spaces. Add pad spaces (or see Program 16) if you want a centered pyramid.
🤔
Did you know?
Outer i runs A.. E. First inner loop prints the left wing (i down to B) by stopping at j > ord('A'). Second inner loop prints A.. i. Because the reverse loop stops before A, the center A is not duplicated.
