Why does the star loop use k from i to rows instead of k from 1 to something?

The loop for (k = i; k <= rows; ++k) runs exactly (rows - i + 1) times, which is the number of stars needed on row i. Equivalently you could use for (k = 1; k <= rows - i + 1; ++k). Starting at i is a common style because it lines up with the column where stars begin after the spaces.

What is the time complexity of this inverted right-aligned triangle star pattern program?

Time complexity is O(n²) where n is the number of rows. Each row i prints (i - 1) spaces and (n - i + 1) stars, which is n characters per row; over n rows the total work is on the order of n².

Inverted Right-Aligned Triangle Star Pattern in C++

Q: How does the program create an inverted right-aligned triangle?

For each row i from 1 to rows, the first inner loop prints (i - 1) spaces. The second inner loop runs k from i to rows and prints a star each time, which prints (rows - i + 1) stars. Row 1 has no spaces and rows stars; each next row adds one space and removes one star, keeping the shape right-aligned and inverted.

Q: What is the difference between this pattern and the standard right-aligned triangle?

Program 3 increases stars per row (1..rows) and uses (rows - i) spaces. Program 4 decreases stars per row using for (k = i; k <= rows) and increases spaces with for (j = 1; j < i; ++j). Same right edge alignment; opposite vertical direction for star counts.

Q: What is the time complexity of this inverted right-aligned triangle star pattern program?

Time complexity is O(n²) where n is the number of rows. Each row i prints (i - 1) spaces and (n - i + 1) stars, which is n characters per row; over n rows the total work is on the order of n².

Beginner

⏱️ 7 min read

📚 Updated: Aug 2025

🎯 2 Code Examples

n rows total

Inverted right-aligned triangle star pattern output in C++ programming

What You'll Learn

This program prints an inverted right-aligned triangle by printing increasing spaces and decreasing stars. The right edge stays aligned.

Row i prints rows - i spaces and i stars (with i decreasing from rows to 1).

⭐ Pattern Output

When you run the program with rows = 5:

Output

*****
 ****
  ***
   **
    *

Complete C++ Program

Fixed rows = 5 version:

C++

#include <iostream>
using namespace std;

int main() {
    int rows = 5;
    int i, j;

    for (i = rows; i >= 1; --i) {
        for (j = 1; j <= rows - i; ++j) cout << " ";
        for (j = 1; j <= i; ++j) cout << "*";
        cout << "\n";
    }

    return 0;
}

🧠 How It Works

Outer loop: `i` from `rows` down

for (i = rows; i >= 1; --i) prints the widest row first. As i shrinks, both the space prefix and star suffix adjust automatically.

Outer

Growing indent

for (j = 1; j <= rows - i; ++j) cout << " "; — when i is large, rows - i is small; as i drops, more leading spaces appear.

Align

Shrinking star block

for (j = 1; j <= i; ++j) cout << "*"; reuses j for the star run. Star count tracks the current i.

Stars

Newline

cout << "\n" after both inner loops. Characters per row: (rows - i) + i = rows, so the right column stays aligned.

Width

Inverted, still right-aligned

O(rows²) writes, O(1) extra space. Wide top rows use the tutorial’s horizontal scroll on phones.

Variation — User Input Version

Accept rows with cin:

C++

#include <iostream>
using namespace std;

int main() {
    int rows;
    int i, j;

    cout << "Enter the number of rows: ";
    cin >> rows;

    for (i = rows; i >= 1; --i) {
        for (j = 1; j <= rows - i; ++j) cout << " ";
        for (j = 1; j <= i; ++j) cout << "*";
        cout << "\n";
    }

    return 0;
}

💡 Tips for Enhancement

Try These

Validate rows > 0 after input
Rewrite the star loop as for (k = 1; k <= rows - i + 1; ++k) and compare readability
Switch to Program 3’s star loop 1..i to get the non-inverted right-aligned triangle
Print a right-aligned number ladder (54321, 5432, …)
Combine with Program 2 mentally: both shrink stars per row; Program 4 adds spaces

Avoid

Using j <= i when you meant j < i for spaces (extra space breaks alignment)
Using k from 1 to i here—that is Program 3’s growing triangle, not this pattern
Forgetting the newline after each row
Assuming cin always succeeds without validating input
Mixing tabs and spaces in the output

Key Takeaways

Row i: print i - 1 spaces, then rows - i + 1 stars.

The star loop k = i .. rows counts exactly those stars in one pass.

Program 3 grows stars; Program 4 shrinks them—same right column, different profile.

Time complexity is O(n²) for n rows.

Master Programs 1–4 and you have left, inverted, right-aligned, and inverted right-aligned triangles.

❓ Frequently Asked Questions

How does the program create an inverted right-aligned triangle?

Print i - 1 spaces, then run k from i to rows for stars. Each row is shorter on the left side of the star block but still ends at the same column, so it looks inverted and right-aligned.

Why does the star loop use k from i to rows?

That range has length rows - i + 1, which is exactly how many stars belong on row i. You can rewrite it as k from 1 to rows - i + 1 if you prefer.

What’s the difference between this pattern and the standard right-aligned triangle?

Program 3 uses spaces rows - i and stars 1..i. Program 4 uses spaces 1..i-1 and stars from i to rows. Same alignment idea; inverted star counts.

What is the time complexity of this program?

O(n²) for n rows: each row does Θ(n) character prints.

Next: Pyramid Pattern

Continue to Program 5 to print a centered pyramid star pattern in C++.

Program 5 →

Did you know?

If you reverse the outer loop of Program 3, you get this inverted right-aligned triangle.

About the author

Mari Selvan M P 🔗

Developer, cloud engineer, and technical writer

Experience 12 years building web and cloud systems
Focus Full Stack Development, AWS, and Developer Education

I write practical tutorials so students and working developers can learn by doing—from databases and APIs to deployment on AWS.

AWS Certified Developer – Associate (Credly)

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