An inverted center-aligned pyramid prints the widest odd-width star row first, then narrows to a single tip star — with leading spaces so each shorter row stays centered.
Remember
Rule: for i from rows down to 1,
print (rows - i) spaces, then (2 * i - 1) stars
*********
*******
*****
***
* ← 5 rows (base on top)
It is the flip of Program 5: keep the same space and star formulas, reverse only the outer loop. That mirrors how Program 2 inverts Program 1, but with odd-width centering.
Approach
How to Solve It
Two ways to emit the same shape — start with nested loops, then optionally shorten with std::string.
Method
Idea
Best for
Nested loops
Countdown i; spaces then odd stars via cout
Learning, interviews, exams
string(n, ch)
Build margin and star run in one call each
Shorter demos once formulas click
Pseudocode
Pseudocode
for i from rows down to 1:
print (rows - i) spaces (no newline)
print (2 * i - 1) stars (no newline)
print newline
Cheat sheet
Goal
Pattern
Countdown rows
for (i = rows; i >= 1; --i)
Leading spaces
for (j = 1; j <= rows - i; ++j) cout << " ";
Odd star run
for (k = 1; k <= 2 * i - 1; ++k) cout << "*";
End the row
cout << "\n";
First printed row
0 spaces + 2 * rows - 1 stars
One-line shortcut
cout << string(rows - i, ' ') << string(2 * i - 1, '*') << "\n";
Print characters without a newline, then end the row once.
Try it
Live Preview
Change the height and the inverted pyramid updates instantly — including the star total (n²).
Whole numbers from 1 to 14. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 25 stars
*********
*******
*****
***
*
Trace
Worked Walkthrough — rows = 4
Trace spaces, stars, and the printed line as i counts down from 4 to 1.
i
Spaces rows - i
Stars 2*i - 1
Printed row
4
0
7
*******
3
1
5
*****
2
2
3
***
1
3
1
*
Star total: 7 + 5 + 3 + 1 = 16 = 4² — same as Program 5, only the print order differs. That square sum is why time is O(n²).
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 — countdown outer loop with space and star inner loops.
C++
#include <iostream>
using namespace std;
int main() {
int rows = 5;
for (int i = rows; i >= 1; --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 from base to tip.
2. Outer loop counts down.i runs from rows down to 1 — widest row first.
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 = 5 you get 0 spaces and 9 stars; when i = 1 you get 4 spaces and 1 star.
Example 2 — User Input Version
Read the height 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 = rows; i >= 1; --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 countdown core. Only the source of rows changes — the space and star 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 margin and odd star run in one call — same shape, no explicit character loops.
C++
#include <iostream>
#include <string>
using namespace std;
int main() {
int rows = 5;
for (int i = rows; i >= 1; --i) {
cout << string(rows - i, ' ')
<< string(2 * i - 1, '*')
<< "\n";
}
return 0;
}
Output
*********
*******
*****
***
*
How It Works
1. Same countdown. Still walk i from rows down to 1.
2. Build each segment.string(rows - i, ' ') is the margin; string(2 * i - 1, '*') is the star run.
3. Print and advance. Append "\n" so the row ends correctly.
Learn the two-loop version first (Examples 1–2) so you can explain both bounds in an interview; treat this as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
++i
Upright pyramid by mistake
If you increment i from 1 to rows, you reprint Program 5. Use --i from rows down to 1.
Tabs
Broken centering
Always print the space character " ", not tabs — tab width varies and skews the tip.
\n inside
Column of stars
If "\n" is inside either inner loop, each character lands on its own line. Print spaces and stars without a newline; end the row only after both loops.
rows = 1
Single tip star
Output is just * — base and tip coincide. A good 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 (Example 3)
O(rows²)
O(rows) per temporary row string
Total stars = 1 + 3 + … + (2n - 1) = n², plus up to Θ(n) spaces per row — still quadratic in n. Same totals as Program 5.
Remember
Key Takeaways
Rule: countdown i; print rows - i spaces then 2 * i - 1 stars.
Flip of Program 5: same formulas — only reverse the outer loop.
Break the row: print "\n" only after both space and star loops.
Complexity:O(n²) time from n² stars; O(1) extra space for nested loops.
One line: for i from rows down to 1, print rows - i spaces and 2 * i - 1 stars, then a newline.
Frequently Asked Questions
The outer loop runs i from rows down to 1. Stars still use 2*i-1, so large i prints many stars first. As i shrinks, stars become 9,7,5,… and spaces (rows-i) grow from 0 upward. Same formulas as Program 5 — only the order of i changes.
Spaces use (rows-i). When i is rows, the margin is 0; when i is 1, the margin is rows-1. As i steps down, the margin grows while 2*i-1 shrinks, which keeps shorter rows centered under the wide top.
Program 5 uses for (i = 1; i <= rows; ++i) so stars grow. Program 6 uses for (i = rows; i >= 1; --i) with the same inner loops, so the base prints first and the tip last.
Program 2 is an inverted left-aligned triangle (i stars, no centering). Program 6 keeps (rows-i) spaces and odd star runs so the tip stays centered.
cout << " " or "*" stays on the same line. cout << "\n" ends the current line. Spaces and stars print without a newline; the row break comes after both inner loops.
O(n²) for n rows. Same totals as Program 5; only iteration order differs. Total stars equal n².
Yes. With the countdown outer loop: 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?
This inverted pyramid is exactly Program 5 with the outer loop reversed — same formulas, different print order. Total stars still equal n².
