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

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

Right-Aligned Triangle Star Pattern in C++

Q: How does the program create right-aligned stars?

The program uses two inner loops after the outer row loop: the first prints (rows - i) space characters to indent the line, and the second prints i stars. Row 1 uses rows-1 spaces and 1 star; row i uses rows-i spaces and i stars, so the stars line up along the right edge.

Q: Why do we need two inner loops instead of one?

One loop prints only one kind of character. Here we need both leading spaces and stars, with different counts per row, so we use one inner loop for spaces (j from 1 to rows - i) and a second for stars (k from 1 to i). Without the space loop, the output would be left-aligned like Program 1.

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

The left-aligned triangle (Program 1) uses one inner loop that prints i stars only. The right-aligned triangle adds a first inner loop that prints (rows - i) spaces before the stars, which shifts each row to the right while still printing i stars per row.

Beginner

⏱️ 7 min read

📚 Updated: Aug 2025

🎯 2 Code Examples

n rows total

Right-aligned triangle star pattern output in C++ programming

What You'll Learn

This program prints a right-aligned triangle by printing spaces first and then stars. Each next row has one fewer leading space and one more star.

Row i prints rows - i spaces and i stars.

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

    return 0;
}

🧠 How It Works

Outer row index

for (i = 1; i <= rows; ++i) walks from the one-star line up to the full-width base. The same j variable is reused for two consecutive inner loops.

Outer

Padding with spaces

for (j = 1; j <= rows - i; ++j) cout << " "; prints the left margin so stars line up on the right.

Align

Star run

A second inner loop reuses j: for (j = 1; j <= i; ++j) cout << "*"; emits i stars after the padding.

Stars

Newline and fixed width

cout << "\n" ends the row. Each line prints (rows - i) + i = rows characters before the newline, so the right edge stays straight.

Width

Right-aligned triangle

Star total is still n(n + 1)/2. O(rows²) output, O(1) extra space. Full-width rows scroll inside the green glyph on narrow viewports.

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 = 1; i <= rows; ++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 cin
Remove the space loop to get the left-aligned triangle from Program 1
Use a forward row loop with (rows - i + 1) stars in one inner loop (same shape, different formulation)
Print digits instead of * for a right-aligned number triangle
Try a hollow right-aligned triangle (border only)

Avoid

Swapping rows - i and i between the two inner loops
Forgetting cout << "\n" or endl after each row
Using j < rows - i when you meant <= (off-by-one spacing)
Mixing tab characters with spaces (alignment breaks in different terminals)
Assuming rows is valid without checking cin / input validation

Key Takeaways

Right alignment = print (rows - i) spaces, then i stars, per row.

Two inner loops are used because spaces and stars have different counts each row.

Star counts per row match Program 1; only the leading padding changes.

Time complexity is O(n²) for n rows (quadratic total output).

This pattern is the usual stepping stone to pyramids and diamonds that mix spaces and stars.

❓ Frequently Asked Questions

How does the program create right-aligned stars?

For row i, print (rows - i) spaces, then i stars. The spaces push the stars to the right so the hypotenuse lines up on the right side of the triangle.

Why do we need two inner loops instead of one?

You need separate loops (or equivalent logic) for two different repeat counts: spaces decrease as i grows, while stars increase. One loop can only drive one of those counts cleanly.

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

Program 1 prints only stars (i per row). Program 3 adds a space loop first so the same i stars appear shifted right.

What is the time complexity of this program?

O(n²) for n rows: each row prints Θ(n) characters, and there are n rows.

Next: Inverted Right-Aligned Triangle

Continue to Program 4 to print the inverted right-aligned star triangle in C++.

Program 4 →

Did you know?

If you reverse the outer loop (start from rows down to 1), you’ll get the inverted right-aligned triangle (Program 4).

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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