Hollow Diamond Star Pattern in C++

Beginner
⏱️ 9 min read
📚 Updated: Aug 2025
🎯 2 Code Examples
2n − 1 rows total
Hollow diamond star pattern output in C++ programming

What You'll Learn

This pattern is Program 7 (inverted V, upper half) plus the lower half from the same diagonal logic as Program 8: after i runs 1…rows, run i from rows - 1 down to 1 with the same inner loops.

Each printed line has width 2 * rows - 1 (9 characters when rows = 5), including trailing spaces after the right edge.

⭐ Pattern Output

When you run the program with rows = 5:

Output
    *    
   * *   
  *   *  
 *     * 
*       *
 *     * 
  *   *  
   * *   
    *
1

Complete C++ Program

Fixed rows = 5 version:

C++
#include <iostream>
using namespace std;

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

    for (i = 1; i <= rows; ++i) {
        for (j = rows; j >= 1; --j) {
            if (i == j) cout << "*";
            else cout << " ";
        }
        for (k = 2; k <= rows; ++k) {
            if (i == k) cout << "*";
            else cout << " ";
        }
        cout << "\n";
    }

    for (i = rows - 1; i >= 1; --i) {
        for (j = rows; j >= 1; --j) {
            if (i == j) cout << "*";
            else cout << " ";
        }
        for (k = 2; k <= rows; ++k) {
            if (i == k) cout << "*";
            else cout << " ";
        }
        cout << "\n";
    }

    return 0;
}

🧠 How It Works

1

Upper half

for (i = 1; i <= rows; ++i) widens the hollow shape to the middle row. Each iteration streams the left j sweep then the right k sweep with cout.

Upper
2

Lower half

for (i = rows - 1; i >= 1; --i) repeats the same inner body so the legs close toward the bottom apex. Starting at rows - 1 skips printing the widest row twice.

Lower
3

Diagonal cells

j runs rows … 1; k runs 2 … rows. Star when i == j or i == k, otherwise cout << " ". Apex rows use only the j side for the single center star.

Edges
4

After every row

cout << "\n" ends each outer iteration in both halves. The inner block is duplicated in source so each phase stays a simple for loop.

Newline
=

Hollow diamond

2 × rows − 1 lines, each 2 × rows − 1 characters wide — O(rows²) output, O(1) extra space. Middle rows scroll horizontally in the preview on small screens.

2

Variation — User Input Version

Accept rows with cin:

C++
#include <iostream>
using namespace std;

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

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

    for (i = 1; i <= rows; ++i) {
        for (j = rows; j >= 1; --j) {
            if (i == j) cout << "*";
            else cout << " ";
        }
        for (k = 2; k <= rows; ++k) {
            if (i == k) cout << "*";
            else cout << " ";
        }
        cout << "\n";
    }

    for (i = rows - 1; i >= 1; --i) {
        for (j = rows; j >= 1; --j) {
            if (i == j) cout << "*";
            else cout << " ";
        }
        for (k = 2; k <= rows; ++k) {
            if (i == k) cout << "*";
            else cout << " ";
        }
        cout << "\n";
    }

    return 0;
}

💡 Tips for Enhancement

Try These

  • Validate rows >= 1 after reading with cin
  • Replace * with #, digits, or custom symbols
  • Print only the first outer loop for an inverted hollow V (Program 7)
  • For a filled diamond, you need rules for the interior (for example distance from the vertical center), not only the two diagonal equalities i == j and i == k
  • Parameterize spacing if you want a wider gap between the two legs

Avoid

  • Starting the second outer loop at i == rows—you would duplicate the widest row
  • Claiming i != j && i != k alone builds a filled diamond (it does not; it inverts the hollow logic incorrectly)
  • Dropping trailing spaces so columns no longer align to width 2 * rows - 1

Key Takeaways

1

Upper half = Program 7; lower half = same inners with i counting down from rows - 1.

2

Widest row appears once, when i == rows in the first loop.

3

Each row: left edge via j, right edge via k; same tests on both halves.

4

Line width is always 2 * rows - 1 characters.

5

Time complexity O(n²) for n = rows.

❓ Frequently Asked Questions

The first outer loop grows i from 1 to rows (hollow inverted V: from the top apex to the wide equator). The second runs i from rows - 1 to 1 (hollow upright V, closing to the bottom apex). Both use the same j / k loops and diagonal conditions.
The widest line is already printed when i == rows in the first loop. Starting the second loop at rows - 1 prevents printing that line twice.
Yes, if you compute an effective row index or branch on upper vs lower half. Two loops are usually clearer and map directly to Programs 7 and 8.
O(n²) for n rows: about 2n - 1 lines, each with two inner passes of length Θ(n).

Next: Filled Diamond Pattern

Continue to Program 10 to print a filled diamond star pattern in C++.

Program 10 →
Did you know?

You can build the hollow diamond by printing Program 7, then printing Program 8 without repeating the widest row (start the lower loop at rows - 1).

About the author

Mari Selvan M P
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)

14 people found this page helpful

C++ Star Pattern 8 C++ Star Pattern 10