It reuses Program 3’s spacing idea, but uses odd star counts so the shape widens on both sides. The same row formula is the upper half of the filled diamond.
Approach
How to Solve It
Two inner loops per row — spaces then odd stars — or the same formulas with std::string.
Method
Idea
Best for
Nested loops
rows - i spaces, then 2 * i - 1 stars
Learning, interviews, exams
string(n, ch)
Build spaces and stars as whole strings
Shorter demos once loops click
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to (rows - i):
print " " (no newline)
for k from 1 to (2 * i - 1):
print "*" (no newline)
print newline
cout << string(rows - i, ' ') << string(2 * i - 1, '*') << "\n";
Printing Stars vs Starting a New Line
API
Effect
Use for
cout << " " / "*"
Stays on the same line
Each space and each *
cout << "\n"
Ends the current line
After spaces and stars for that row
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the pyramid updates instantly — including the perfect-square star total.
Whole numbers from 1 to 15. Base width = 2 * rows - 1; total stars = rows².
Live result5 rows · 25 stars
*
***
*****
*******
*********
Trace
Worked Walkthrough — rows = 4
Trace spaces and odd star counts for each outer-loop value of i.
i
Spaces rows - i
Stars 2 * i - 1
Printed row
1
3
1
*
2
2
3
***
3
1
5
*****
4
0
7
*******
Total stars: 1 + 3 + 5 + 7 = 16 = 4². Base width: 2×4 - 1 = 7.
Code
C++ Programs
Three complete programs: fixed rows, console input, and a string shortcut. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — ideal for first demos and screenshots.
C++
#include <iostream>
using namespace std;
int main() {
int rows = 5;
for (int i = 1; i <= rows; ++i) {
for (int j = 1; j <= rows - i; ++j) {
cout << " ";
}
for (int k = 1; k <= 2 * i - 1; ++k) {
cout << "*";
}
cout << "\n";
}
return 0;
}
Output
*
***
*****
*******
*********
How It Works
1. Set height.rows = 5 means five lines; the base has 2 * 5 - 1 = 9 stars.
2. Outer loop picks the row.i runs from 1 to rows.
3. Spaces then stars. Print rows - i spaces, then 2 * i - 1 stars via cout (no newline yet).
4. Break the line.cout << "\n" after both inner loops starts the next row.
When i = 1: 4 spaces + 1 star. When i = 5: 0 spaces + 9 stars.
Example 2 — User Input Version
Read the row count at runtime. Prefer checking cin for failure in real apps (shown in the tip below).
C++
#include <iostream>
using namespace std;
int main() {
int rows;
cout << "Enter the number of rows: ";
cin >> rows;
for (int i = 1; i <= rows; ++i) {
for (int j = 1; j <= rows - i; ++j) {
cout << " ";
}
for (int k = 1; k <= 2 * i - 1; ++k) {
cout << "*";
}
cout << "\n";
}
return 0;
}
Output (when user enters 4)
Enter the number of rows: 4
*
***
*****
*******
How It Works
1. Prompt and read. Ask for a row count, then cin >> rows stores the integer.
2. Same nested-loop core. Only the source of rows changes — the print logic matches Example 1.
3. Safer input tip. Letters leave cin in a failed state. Prefer:
Safer input
if (!(cin >> rows) || rows < 1) {
cout << "Enter a positive whole number.\n";
return 1;
}
Example 3 — string for Spaces and Stars
Build each row in one statement — same shape, no explicit space/star character loops.
C++
#include <iostream>
#include <string>
using namespace std;
int main() {
int rows = 5;
for (int i = 1; i <= rows; ++i) {
cout << string(rows - i, ' ')
<< string(2 * i - 1, '*')
<< "\n";
}
return 0;
}
Output
*
***
*****
*******
*********
How It Works
1. One outer loop. Still walk i from 1 to rows.
2. Build the row.string(rows - i, ' ') for padding; string(2 * i - 1, '*') for the odd star run; then "\n".
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.
2 * i
Even width
Use 2 * i - 1 (odd). Even widths lose the classic single-center peak.
j < rows - i
One space short
Space loop must be j <= rows - i. A strict < shifts the peak.
Only i stars
Program 3 shape
Same spaces with i stars is the right-aligned triangle, not a full pyramid.
rows = 1
Single star
0 spaces + 1 star — a good tip sanity check.
rows ≤ 0
Empty output
Outer loop never runs. Validate and re-prompt for interactive programs.
Bad input
Check cin
Letters fail cin >> rows — test with if (!(cin >> rows)).
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(rows²)
O(1)
string shortcut (Example 3)
O(rows²)
O(rows) per temporary row string
Each of n rows prints Θ(n) characters (spaces + stars). Total stars = 1 + 3 + … + (2n - 1) = n².
Remember
Key Takeaways
Formulas:rows - i spaces and 2 * i - 1 stars.
Odd widths: keep 2 * i - 1 for a single centered peak.
Break the row:cout characters without newline; then cout << "\n".
Complexity:O(n²) time; total stars = n²; O(1) extra space for nested loops.
One line: print rows - i spaces, then 2 * i - 1 stars — that is the centered pyramid.
Frequently Asked Questions
2*i-1 gives odd lengths 1, 3, 5, … so each row adds one star on both sides. Using only i stars per row would not form the usual symmetric centered pyramid.
Printing (rows - i) spaces before the stars shifts the star block left as i grows, keeping the peak centered when the font is fixed-width.
Yes. Keep the same inner loops but run the outer loop from rows down to 1. The first printed row is the widest; later rows narrow toward the tip. See Program 6.
The last row has 2 * rows - 1 stars and no leading spaces when i equals rows.
O(n²) for n rows. Each row prints Theta(n) characters in the worst case; there are n rows. Total stars equal n².
Program 3 uses the same (rows - i) spaces but only i stars. Program 5 uses 2*i-1 stars so the shape widens on both sides.
Yes. cout << string(rows - i, ' ') << string(2 * i - 1, '*') << "\n".
After cin >> rows, check if (!(cin >> rows)) or cin.fail() so bad input does not leave rows unset.
🤔
Did you know?
Odd star counts 1, 3, 5, … come from 2 * i - 1. Their sum for n rows is n² — so total stars grow as a perfect square. This pyramid is also the upper half of the filled diamond.
