C Program to Print Diamond Pattern

A C program to print a diamond pattern uses nested for loops to print spaces and stars in increasing and then decreasing order, like for (int i = 1; i <= n; i++) for the top and for (int i = n-1; i > 0; i--) for the bottom part.

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 and decreasing spaces before them. 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

c

#include <stdio.h>

int main() {
    int n = 5;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n - i; j++) printf(" ");
        for (int k = 1; k <= 2 * i - 1; k++) printf("*");
        printf("\n");
    }
    for (int i = n - 1; i > 0; i--) {
        for (int j = 1; j <= n - i; j++) printf(" ");
        for (int k = 1; k <= 2 * i - 1; k++) printf("*");
        printf("\n");
    }
    return 0;
}

Output

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

Dry Run

Let's trace 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 top half

The first loop prints lines with spaces decreasing and stars increasing using n - i spaces and 2*i - 1 stars.

Step 2: Print bottom half

The second loop prints lines with spaces increasing and stars decreasing, mirroring the top half.

Step 3: Use nested loops

Nested loops control spaces and stars separately for each line to form the diamond shape.

Alternative Approaches

Using while loops

c

#include <stdio.h>

int main() {
    int n = 5, i = 1, j, k;
    while (i <= n) {
        j = 1;
        while (j <= n - i) { printf(" "); j++; }
        k = 1;
        while (k <= 2 * i - 1) { printf("*"); k++; }
        printf("\n");
        i++;
    }
    i = n - 1;
    while (i > 0) {
        j = 1;
        while (j <= n - i) { printf(" "); j++; }
        k = 1;
        while (k <= 2 * i - 1) { printf("*"); k++; }
        printf("\n");
        i--;
    }
    return 0;
}

Uses while loops instead of for loops; slightly longer but same logic.

Using a single loop with condition

c

#include <stdio.h>

int main() {
    int n = 5, total = 2 * n - 1;
    for (int line = 1; line <= total; line++) {
        int stars = line <= n ? 2 * line - 1 : 2 * (total - line + 1) - 1;
        int spaces = (total - stars) / 2;
        for (int s = 0; s < spaces; s++) printf(" ");
        for (int st = 0; st < stars; st++) printf("*");
        printf("\n");
    }
    return 0;
}

Uses one loop for both halves, calculating stars and spaces per line.

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 n times, resulting in O(n^2) time.

Space Complexity

Only a few variables are used; no extra memory proportional to input size, so space complexity is O(1).

Which Approach is Fastest?

All approaches have similar O(n^2) time; using a single loop with calculations can be slightly cleaner but not faster.

Approach	Time	Space	Best For
Nested for loops	O(n^2)	O(1)	Simple and clear code
While loops	O(n^2)	O(1)	Alternative syntax preference
Single loop with condition	O(n^2)	O(1)	Compact code combining halves