C++ Program to Print Diamond Pattern

You can print a diamond pattern in C++ using nested for loops to print spaces and stars in the right order; for example, use for (int i = 1; i <= n; i++) to print the top half and for (int i = n-1; i >= 1; i--) for the bottom half.

Examples

Inputn = 1

Output*

Inputn = 3

Output * *** ***** *** *

Inputn = 5

Output * *** ***** ******* ********* ******* ***** *** *

How to Think About It

To print a diamond pattern, first print the top half by increasing the number of stars each line while decreasing spaces, then print the bottom half by decreasing stars and increasing spaces. Use loops to control spaces and stars for each line.

Algorithm

1

Get input number n for the diamond's half height

2

For each line i from 1 to n, print (n - i) spaces and (2 * i - 1) stars

3

For each line i from n-1 down to 1, print (n - i) spaces and (2 * i - 1) stars

4

End program

Code

cpp

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n - i; j++) cout << ' ';
        for (int j = 1; j <= 2 * i - 1; j++) cout << '*';
        cout << '\n';
    }
    for (int i = n - 1; i >= 1; i--) {
        for (int j = 1; j <= n - i; j++) cout << ' ';
        for (int j = 1; j <= 2 * i - 1; j++) cout << '*';
        cout << '\n';
    }
    return 0;
}

Output

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

Dry Run

Let's trace the diamond pattern for n=3 through the code

1

Top half line 1

Print 2 spaces and 1 star: ' *'

2

Top half line 2

Print 1 space and 3 stars: ' ***'

3

Top half line 3

Print 0 spaces and 5 stars: '*****'

4

Bottom half line 1

Print 1 space and 3 stars: ' ***'

5

Bottom half line 2

Print 2 spaces and 1 star: ' *'

Line	Spaces	Stars	Output
1	2	1	*
2	1	3	***
3	0	5	*****
4	1	3	***
5	2	1	*

Why This Works

Step 1: Print spaces first

Each line starts with spaces to center the stars; the number of spaces is n - i for line i.

Step 2: Print stars next

Stars increase by two each line in the top half using 2 * i - 1 to form the diamond shape.

Step 3: Repeat for bottom half

The bottom half reverses the process, decreasing stars and increasing spaces to complete the diamond.

Alternative Approaches

Using while loops

cpp

#include <iostream>
using namespace std;

int main() {
    int n, i = 1;
    cin >> n;
    while (i <= n) {
        int j = 1;
        while (j <= n - i) { cout << ' '; j++; }
        j = 1;
        while (j <= 2 * i - 1) { cout << '*'; j++; }
        cout << '\n';
        i++;
    }
    i = n - 1;
    while (i > 0) {
        int j = 1;
        while (j <= n - i) { cout << ' '; j++; }
        j = 1;
        while (j <= 2 * i - 1) { cout << '*'; j++; }
        cout << '\n';
        i--;
    }
    return 0;
}

This uses while loops instead of for loops; it is more verbose but works the same.

Using a single loop with conditions

cpp

#include <iostream>
using namespace std;

int main() {
    int n;
    cin >> n;
    int total = 2 * n - 1;
    for (int i = 1; i <= total; i++) {
        int stars = i <= n ? 2 * i - 1 : 2 * (total - i + 1) - 1;
        int spaces = (total - stars) / 2;
        for (int j = 0; j < spaces; j++) cout << ' ';
        for (int j = 0; j < stars; j++) cout << '*';
        cout << '\n';
    }
    return 0;
}

This uses one loop for both halves by calculating stars and spaces based on line number.

Complexity: O(n^2) time, O(1) space

Time Complexity

The program uses nested loops where the outer loop runs about 2n times and inner loops run up to 2n times, resulting in O(n^2) time.

Space Complexity

Only a few variables are used; no extra data structures, so space complexity is O(1).

Which Approach is Fastest?

All approaches have similar O(n^2) time; using a single loop can be slightly cleaner but does not improve performance.

Approach	Time	Space	Best For
Two for loops (top and bottom)	O(n^2)	O(1)	Simple and clear code
While loops	O(n^2)	O(1)	Alternative loop style
Single loop with conditions	O(n^2)	O(1)	Compact code combining halves