A centered number diamond prints ascending digit sequences on each row — growing to a peak of odd length, then mirroring back down — with leading spaces to keep the shape centered.
Remember
Rule: row i prints 1..(2*i-1); top then bottom half
1
123
12345
1234567
123456789
1234567
12345
123
1 ← levels = 5
Unlike Program 43 (right-aligned triangle only), this shape has two phases: grow to the peak, then shrink symmetrically.
Approach
How to Solve It
Run a top half that grows, then a bottom half that shrinks — each row: spaces, then digits 1..(2*i-1).
Method
Idea
Best for
Two outer loops
Top 1..levels, bottom levels-1..1
Learning, interviews, exams
Spaces + digits
Indent, then k = 1 .. 2*i-1
Centered diamond demos
Pseudocode
Pseudocode
for i from 1 to levels: // top
print (levels - i) spaces
for k from 1 to (2*i - 1):
print k (no newline)
print newline
for i from levels-1 down to 1: // bottom
print (levels - i) spaces
for k from 1 to (2*i - 1):
print k (no newline)
print newline
Cheat sheet
Goal
Pattern
Top half
for (i = 1; i <= levels; i++)
Bottom half
for (i = levels - 1; i >= 1; i--)
Top spaces
for (j = i; j < levels; j++) cout << " ";
Bottom spaces
for (j = levels; j > i; j--) cout << " ";
Digits on row i
for (k = 1; k < i * 2; k++) cout << k;
End the row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << " " / cout << k
Stays on the same line
Each space or digit
cout << "\n"
Ends the current line
After the digit loop
Print spaces and digits without a newline, then end the row once. cout << endl also ends the line and flushes; "\n" is enough for these demos.
Try it
Live Preview
Change the level count and the centered diamond updates instantly — capped at 5 so peak digits stay single-digit (1..9).
Whole numbers from 1 to 5. Tap a chip or type a value — the preview redraws as you go.
Live result5 levels · 41 digits
1
123
12345
1234567
123456789
1234567
12345
123
1
Trace
Worked Walkthrough — levels = 3
Trace each half: spaces, digit count 2*i-1, and the printed row.
Half
i
Spaces
Digits
Printed row
Top
1
2
1
1
Top
2
1
123
123
Top
3
0
12345
12345
Bottom
2
1
123
123
Bottom
1
2
1
1
Peak row is printed once in the top half — the bottom starts at levels - 1 so it is not duplicated. Digit total for n levels = n² + (n-1)².
Code
C++ Programs
Three complete programs: fixed levels, cin input, and a compact trace demo. Use View Output to reveal sample results.
Example 1 — Fixed levels = 5
Hard-coded level count — ideal for first demos and screenshots.
C++
#include <iostream>
using namespace std;
int main()
{
int i, j, k;
for (i = 1; i <= 5; i++)
{
for (j = i; j < 5; j++)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
for (i = 4; i >= 1; i--)
{
for (j = 5; j > i; j--)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
return 0;
}
Output
1
123
12345
1234567
123456789
1234567
12345
123
1
How It Works
1. Top half grows.i runs 1..5; spaces shrink while digits grow to 2*i-1.
2. Digit loop.for (k = 1; k < i * 2; k++) prints exactly 2*i-1 digits — e.g. i = 3 prints 12345.
3. Bottom half mirrors.i runs 4..1 so the peak is not printed twice; spaces grow again as rows shrink.
Example 2 — User Input Levels
Read the level count at runtime and use it in both halves. Prefer validating cin (tip below).
C++
#include <iostream>
using namespace std;
int main()
{
int levels;
int i, j, k;
cout << "Enter levels: ";
cin >> levels;
for (i = 1; i <= levels; i++)
{
for (j = i; j < levels; j++)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
for (i = levels - 1; i >= 1; i--)
{
for (j = levels; j > i; j--)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
return 0;
}
Output (when user enters 3)
Enter levels: 3
1
123
12345
123
1
How It Works
1. Prompt and read. Ask for a level count, then store it in levels.
2. Same diamond core. Only the source of levels changes — top and bottom logic match Example 1.
3. Safer input tip. Prefer:
Safer input
if (!(cin >> levels) || levels < 1)
{
cout << "Enter a whole number of levels (1 or more).\n";
return 1;
}
Example 3 — Compact levels = 3
Same space and number loops with a smaller level count for quick tracing on paper.
C++
#include <iostream>
using namespace std;
int main()
{
int levels = 3;
int i, j, k;
for (i = 1; i <= levels; i++)
{
for (j = i; j < levels; j++)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
for (i = levels - 1; i >= 1; i--)
{
for (j = levels; j > i; j--)
cout << " ";
for (k = 1; k < i * 2; k++)
cout << k;
cout << "\n";
}
return 0;
}
Output
1
123
12345
123
1
How It Works
1. Only levels changes. The space and digit loops stay identical to Examples 1 and 2.
2. Trace on paper. Row lengths are 1, 3, 5, then 3, 1 — easy to check by hand before scaling up.
3. Peak once. Bottom starts at levels - 1, so 12345 appears only once.
Edge Cases & Pitfalls
Check these before calling the solution done.
peak twice
Duplicated middle
If the bottom half starts at levels instead of levels - 1, the widest row prints twice. Start at levels - 1.
no spaces
Left-snapped diamond
Without the space loops, rows grow and shrink flush left. Keep the indent loops for centering.
k <= i
Wrong digit count
Stopping at i prints only i digits. Use k < i * 2 for odd lengths 1, 3, 5, ….
\n early
Column of digits
If cout << "\n" is inside the digit loop, each digit lands on its own line. End the row only after the digit loop.
levels > 5
Multi-digit values
For i > 5, k reaches 10+. Cap at 5 for a clean single-digit diamond, or format with spaces/setw.
Bad cin
Validate levels
Check cin >> levels and require levels >= 1 before the loops.
Analysis
Time and Space Complexity
Program
Time
Extra space
Top + bottom halves (Examples 1–3)
O(levels²)
O(1)
Top digits = n²; bottom digits = (n-1)²; total = n² + (n-1)² — still quadratic in n. Only a few integers of extra memory.
Remember
Key Takeaways
Rule: row i prints 1..(2*i-1); grow with top half, shrink with bottom half.
Peak once: start the bottom half at levels - 1 so the widest row is not duplicated.
Break the row: call cout << "\n" only after the digit loop.
Complexity:O(n²) time from the digit totals; O(1) extra space.
One line: for each half, print spaces then digits 1..(2*i-1), grow to the peak, then mirror down without reprinting it.
Frequently Asked Questions
A centered diamond of ascending digit sequences: 1, 123, 12345, 1234567, 123456789, then mirrors back down to 1.
The row prints 2*i-1 digits (odd length): 1, 3, 5, 7, 9, ... using for (k = 1; k < i*2; k++).
A space loop prints leading spaces before the numbers. Top half uses for (j = i; j < levels; j++); bottom half uses for (j = levels; j > i; j--).
The first loop builds the top half (i = 1..levels). The second mirrors back down (i = levels-1..1) to complete the diamond.
Program 43 is a right-aligned triangle with fixed-width columns. Program 44 is a symmetric centered diamond with odd-length digit rows.
Printing a digit stays on the same line. Printing a newline ends the current line. Digits use cout without a newline; the row break uses cout << "\n" after the number loop.
O(n²) for n levels because each level prints O(n) digits and there are O(n) levels.
After cin >> levels, check failure and require levels ≥ 1 before the loops. Cap at 5 if you want single-digit values only.
🤔
Did you know?
This pattern prints a top half (1..levels) and a bottom half (levels-1..1). Each row prints 2*i-1 digits (1 to 2*i-1) with leading spaces to center the diamond.
