A reverse centered alphabet pyramid prints Program 28’s layered rows down to the A-center line, then mirrors them back up to the top letter — without printing the center twice.
Remember
Rule: PrintRow(i) = left k..0 + right 1..k with (j > i ? j : i)
Phase 1: i = k..0 Phase 2: i = 1..k
E E E E E E E E E
E D D D D D D D E
E D C C C C C D E
E D C B B B C D E
E D C B A B C D E ← center once
E D C B B B C D E
E D C C C C C D E
E D D D D D D D E
E E E E E E E E E ← top = 'E' (9×9)
Program 28 is exactly the upper half. The lower phase starts at i = 1 (B) so the A-center row is not duplicated.
Approach
How to Solve It
Two ways to emit the same diamond — duplicate the row body in both phases, or factor helpers.
Method
Idea
Best for
Two phases inline
Upper k..0, lower 1..k, same row body
Learning, interviews, exams
printCell + printRow
One place owns the floor rule and both halves
Cleaner demos once the rule clicks
Pseudocode
Pseudocode
k = top - 'A'
define PrintRow(i):
for j from k down to 0: print (j > i ? alpha[j] : alpha[i])
for j from 1 to k: print (j > i ? alpha[j] : alpha[i])
print newline
for i from k down to 0: PrintRow(i) // upper
for i from 1 to k: PrintRow(i) // lower
Print cells without a newline, then end the row once.
Try it
Live Preview
Change the top letter and the reverse-centered diamond updates instantly — rows = width = 2k + 1.
One letter from A to F. Tap a chip or type a letter — the preview redraws as you go.
Live resultTop E · 9 rows · width 9
E E E E E E E E E
E D D D D D D D E
E D C C C C C D E
E D C B B B C D E
E D C B A B C D E
E D C B B B C D E
E D C C C C C D E
E D D D D D D D E
E E E E E E E E E
Trace
Worked Walkthrough — Top = E (k = 4)
Trace each floor and the resulting 9-letter line across both phases.
Phase
i
Floor
Printed row
Upper
4
E
E E E E E E E E E
Upper
3
D
E D D D D D D D E
Upper
2
C
E D C C C C C D E
Upper
1
B
E D C B B B C D E
Upper
0
A
E D C B A B C D E
Lower
1
B
E D C B B B C D E
Lower
2
C
E D C C C C C D E
Lower
3
D
E D D D D D D D E
Lower
4
E
E E E E E E E E E
Width is always 2×4 + 1 = 9. Total rows are 2×5 − 1 = 9. The A-center row appears only once (upper phase).
Code
C++ Programs
Three complete programs: fixed A–E, top-letter input, and a helper-function rewrite. Use View Output to reveal sample results.
Example 1 — Fixed A–E
Same row logic as Program 28, printed in two phases to complete the reverse centered pyramid.
C++
#include <iostream>
using namespace std;
int main()
{
int k = 4; // index for 'E'
const char* alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
// Upper half (E down to A)
for (int i = k; i >= 0; i--)
{
for (int j = k; j >= 0; j--)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
for (int j = 1; j <= k; j++)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
cout << "\n";
}
// Lower half (B up to E) — skip repeating the A row
for (int i = 1; i <= k; i++)
{
for (int j = k; j >= 0; j--)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
for (int j = 1; j <= k; j++)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
cout << "\n";
}
return 0;
}
Output
E E E E E E E E E
E D D D D D D D E
E D C C C C C D E
E D C B B B C D E
E D C B A B C D E
E D C B B B C D E
E D C C C C C D E
E D D D D D D D E
E E E E E E E E E
How It Works
1. Fix the top index.k = 4 means the highest letter is E.
2. Upper phase. Floor i runs from k down to 0 — same as Program 28 through the A-center row.
3. Lower phase. Floor i runs from 1 to k so the center line is not printed twice.
4. Same row body. Both phases use left k..0 and right 1..k with j > i ? alpha[j] : alpha[i].
Example 2 — Top Letter Input
Works for A..top with the same two-phase pyramid. Prefer validating a single A–Z character in real apps.
C++
#include <iostream>
using namespace std;
int main()
{
char top;
cout << "Enter top letter (like E): ";
cin >> top;
int k = top - 'A';
const char* alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = k; i >= 0; i--)
{
for (int j = k; j >= 0; j--)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
for (int j = 1; j <= k; j++)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
cout << "\n";
}
for (int i = 1; i <= k; i++)
{
for (int j = k; j >= 0; j--)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
for (int j = 1; j <= k; j++)
cout << (j > i ? alpha[j] : alpha[i]) << " ";
cout << "\n";
}
return 0;
}
Output (when user enters C)
Enter top letter (like E): C
C C C C C
C B B B C
C B A B C
C B B B C
C C C C C
How It Works
1. Prompt and scale.k = top - 'A' sets both phases and both halves.
2. Same diamond core. For top = C you get 5 rows of width 5 (2k + 1).
3. Safer input tip. Prefer:
Safer input
#include <cctype>
if (!(cin >> top))
{
cout << "Enter one letter from A to Z.\n";
return 1;
}
top = (char)toupper((unsigned char)top);
if (top < 'A' || top > 'Z')
{
cout << "Enter one letter from A to Z.\n";
return 1;
}
Example 3 — Helper Functions
Often clearer: one function owns the floor rule; another prints a full row so both phases stay thin.
C++
#include <iostream>
using namespace std;
void printCell(const char* alpha, int j, int i)
{
cout << (j > i ? alpha[j] : alpha[i]) << " ";
}
void printRow(const char* alpha, int k, int i)
{
for (int j = k; j >= 0; j--)
printCell(alpha, j, i);
for (int j = 1; j <= k; j++)
printCell(alpha, j, i);
cout << "\n";
}
int main()
{
int k = 4;
const char* alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (int i = k; i >= 0; i--)
printRow(alpha, k, i);
for (int i = 1; i <= k; i++)
printRow(alpha, k, i);
return 0;
}
Output
E E E E E E E E E
E D D D D D D D E
E D C C C C C D E
E D C B B B C D E
E D C B A B C D E
E D C B B B C D E
E D C C C C C D E
E D D D D D D D E
E E E E E E E E E
How It Works
1. printCell owns the rule.j > i lives in one place.
2. printRow owns both halves. Left k..0 and right 1..k, then cout << "\n".
3. Thin phases. The two outer loops only decide which floors to visit.
Edge Cases & Pitfalls
Check these before calling the solution done.
i = 0
Duplicate center row
If the lower phase starts at 0, the A-center line prints twice. Keep for (i = 1; i <= k; i++).
Right j = 0
Duplicate center A in a row
The right half of each row must start at 1, same as Program 28.
\n early
Column of letters
If cout << "\n" is inside either half-loop, each cell lands on its own line. Print cells without a newline; end the row only after both halves.
top = A
Single A
When k = 0, upper prints A and lower never runs. A good sanity check.
Upper only
Program 28 by mistake
Forgetting the lower phase leaves the open square. Add for (i = 1; i <= k; i++) with the same row printer.
Bad cin
Validate one letter
Check cin >> top, optionally uppercase, and require A–Z before the outer loops.
Analysis
Time and Space Complexity
Program
Time
Extra space
Two phases inline (Examples 1–2)
O(n²)
O(1) (plus the fixed alphabet table)
Helper functions (Example 3)
O(n²)
O(1)
For n = k + 1 letters, there are 2n - 1 rows of width 2n - 1 — still quadratic in n. Roughly twice Program 28’s cell count, minus one shared center row.
Remember
Key Takeaways
Rule: same printRow as Program 28 — upper k..0, lower 1..k.
No double center: lower phase starts at B (i = 1).
Break the row: call cout << "\n" only after both half-loops.
Complexity:O(n²) time; O(1) extra space.
One line: print Program 28’s rows from k down to 0, then again from 1 up to k.
Frequently Asked Questions
The first loop decreases i from E down to A, printing each layered row toward the center. The second increases i from B back to E with the same row rule so the pyramid widens again without repeating the A-centered row.
Because the A-centered row already appears in the upper half. Starting from B prevents duplicating the center line.
If n is the number of letters from A to the top letter, total rows are 2n−1 (9 rows for A..E).
It prints the border letter when the column index j is above the current row floor i; otherwise it prints the floor letter. The same rule applies on both left and right halves of every row.
The left scan goes E down to A; the right scan goes B up to E so the middle A appears once and the row mirrors.
O(n²) because there are O(n) rows and each row prints O(n) cells.
Use cin >> top, require A–Z (optionally toupper), and reject failed input.
Program 28 is exactly the upper half of this pyramid. Program 29 reuses that row logic, then mirrors upward from B to E for the closed diamond.
🤔
Did you know?
Reuse Program 28’s row logic twice: first with i from E down to A, then with i from B up to E so the center row is not duplicated. Each row stays full width (2n-1 cells); total rows are also 2n-1 for n letters.
