A number-star diamond grows from 1 to n*n*…*n, then mirrors back down — each row alternates the row number and *.
Remember
Rule: top i = 1..n, bottom i = n-1..1
for j from 1 to 2*i-1:
odd j → print i even j → print *
1
2*2
3*3*3
4*4*4*4
5*5*5*5*5
4*4*4*4
3*3*3
2*2
1 ← n = 5
Follows the right-aligned descending triangle in Program 30; next is the increasing triangle from 11 in Program 32.
Approach
How to Solve It
Two outer loops build the diamond. One shared inner rule: print 2*i−1 characters, alternating digit and star with j % 2.
Method
Idea
Best for
Two halves + modulus
Top 1..n, bottom n-1..1; even j → *
Learning, interviews, exams
Ternary form
(j % 2 == 0) ? cout << "*" : cout << i
Shorter demos once if/else is clear
Pseudocode
Pseudocode
for i from 1 to n:
for j from 1 to 2*i - 1:
if j % 2 == 0: print *
else: print i
print newline
for i from n - 1 down to 1:
for j from 1 to 2*i - 1:
if j % 2 == 0: print *
else: print i
print newline
Cheat sheet
Goal
Pattern
Top half
for (i = 1; i <= n; i++)
Bottom half
for (i = n - 1; i >= 1; i--)
Row length
for (j = 1; j < i * 2; j++) → 2*i-1 chars
Alternate
if (j % 2 == 0) cout << "*"; else cout << i;
End of row
cout << "\n";
Printing Numbers vs Starting a New Line
API
Effect
Use for
cout << i / "*"
Stays on the same line
Each digit or star on the row
cout << "\n"
Ends the current line
After the inner loop finishes
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the height and the number-star diamond updates instantly — capped at 7 for readable demos.
Whole numbers from 1 to 7. Tap a chip or type a value — the preview redraws as you go.
Live resultn = 5 · 41 chars
1
2*2
3*3*3
4*4*4*4
5*5*5*5*5
4*4*4*4
3*3*3
2*2
1
Trace
Worked Walkthrough — Top Half n = 5
Trace how each top-half row builds 2*i−1 characters. The bottom half repeats rows 4..1.
i
Chars
Pattern
Prints
1
1
i
1
2
3
i * i
2*2
3
5
i * i * i
3*3*3
4
7
i * … * i
4*4*4*4
5
9
peak
5*5*5*5*5
Top half = n² chars; bottom = (n−1)². Total lines = 2n − 1 → O(n²).
Code
C++ Programs
Three complete programs: fixed n = 5, cin input with ternary form, and a compact n = 3 demo. Use View Output to reveal sample results.
Example 1 — Fixed n = 5
Hard-coded height — top half grows, bottom half mirrors, if/else on j % 2.
C++
#include <iostream>
using namespace std;
int main()
{
int i, j;
for (i = 1; i <= 5; i++)
{
for (j = 1; j < i * 2; j++)
{
if (j % 2 == 0)
cout << "*";
else
cout << i;
}
cout << "\n";
}
for (i = 4; i >= 1; i--)
{
for (j = 1; j < i * 2; j++)
{
if (j % 2 == 0)
cout << "*";
else
cout << i;
}
cout << "\n";
}
return 0;
}
Output
1
2*2
3*3*3
4*4*4*4
5*5*5*5*5
4*4*4*4
3*3*3
2*2
1
How It Works
1. Top half grows.i from 1 to 5 builds longer rows up to the peak.
2. Modulus alternates. Odd j prints i; even j prints *.
3. Bottom mirrors. Second loop starts at n - 1 so the peak is not printed twice.
Example 2 — User Input Height
Read n with cin, validate, then use ternary conditions for each position.
C++
#include <iostream>
using namespace std;
int main()
{
int n;
int i, j;
cout << "Enter n: ";
if (!(cin >> n) || n <= 0)
{
cout << "Please enter a positive integer.\n";
return 1;
}
for (i = 1; i <= n; i++)
{
for (j = 1; j < i * 2; j++)
(j % 2 == 0) ? cout << "*" : cout << i;
cout << "\n";
}
for (i = n - 1; i >= 1; i--)
{
for (j = 1; j < i * 2; j++)
(j % 2 == 0) ? cout << "*" : cout << i;
cout << "\n";
}
return 0;
}
Output (when user enters 3)
Enter n: 3
1
2*2
3*3*3
2*2
1
How It Works
1. Prompt and validate. Reject failed reads and non-positive values before printing.
2. Same core. Ternary form matches Example 1’s if/else — only n comes from the user.
3. Safer input tip. Cap demos for readable output:
Safer input
if (!(cin >> n) || n < 1 || n > 7)
{
cout << "Enter a whole number from 1 to 7.\n";
return 1;
}
Example 3 — Compact n = 3
Same structure with a small peak — easy to confirm the bottom starts at n - 1.
C++
#include <iostream>
using namespace std;
int main()
{
int n = 3;
int i, j;
for (i = 1; i <= n; i++)
{
for (j = 1; j < i * 2; j++)
{
if (j % 2 == 0)
cout << "*";
else
cout << i;
}
cout << "\n";
}
for (i = n - 1; i >= 1; i--)
{
for (j = 1; j < i * 2; j++)
{
if (j % 2 == 0)
cout << "*";
else
cout << i;
}
cout << "\n";
}
return 0;
}
Output
1
2*2
3*3*3
2*2
1
How It Works
1. Five lines. Peak 3*3*3 appears once; bottom starts at i = 2.
2. Trace on paper. If the bottom loop starts at n, the peak prints twice.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for height 5 or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
i = n
Doubled peak
Starting the bottom loop at n reprints the widest row. Use i = n - 1.
j <=
Wrong row length
j <= i * 2 adds an extra character. Keep j < i * 2 for exactly 2*i-1 chars.
% flip
Stars in wrong spots
Even j must print * and odd j must print i — flipping them breaks the pattern.
\n inside
Broken rows
If cout << "\n" sits inside the inner loop, you get one character per line. Call it only after the loop.
n = 1
Single digit
Output is just 1 — the bottom loop never runs. A good sanity check for input validation.
cin
Check the stream
Validate cin >> n before looping — a failed read leaves n unset or stale.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(n²)
O(1)
Compact n = 3 (Example 3)
O(n²)
O(1)
Top half prints n² characters; bottom prints (n−1)². Total ≈ 2n² → O(n²) time. Only a few loop variables are needed.
Remember
Key Takeaways
Two halves: top 1..n, bottom n-1..1 so the peak appears once.
Modulus rule: odd j → digit i; even j → *.
Break the row: characters in the inner loop; cout << "\n" after it finishes.
Complexity:O(n²) time from both halves; O(1) extra space.
One line: for each half, print 2*i−1 chars alternating i and *, then newline.
Frequently Asked Questions
A number-star diamond: for n=5 you get 1 / 2*2 / 3*3*3 / 4*4*4*4 / 5*5*5*5*5, then the same rows mirrored back down to 1.
The inner loop runs while j < i*2, which prints 1, 3, 5, 7, 9 characters for i = 1..5.
It checks j % 2. Even j prints '*', odd j prints the current row number i.
The first loop builds the top half (i = 1..n). The second mirrors back down (i = n-1..1) to complete the diamond.
Program 30 is a right-aligned descending triangle. Program 31 alternates digits and stars in a symmetric diamond shape.
cout << i (or "*") stays on the same line. cout << "\n" ends the row after the inner loop finishes.
After prompting, use if (!(cin >> n) || n <= 0) to reject bad input before printing.
O(n²) for height n because total printed characters grow as n² + (n-1)² across both halves.
🤔
Did you know?
This pattern prints a top half (1..n) and a bottom half (n-1..1). Each row prints 2*i-1 characters, alternating the row number and * using j % 2.
