Question

How do I make this:

*******
-*****-
--***--
---*---
--***--
-*****-
*******

The following is my code that I have written to try to accomplish the above, but it is not working as expected:

    public static void stars(/*int jmlBaris*/) {
    for ( int i = 7; i >= 1; i-=2) {
        for (int j = 1; j <= i; j++) {

            System.out.print("*");
        }
        System.out.println("");
    }

    for (int i = 1; i <= 7; i+=2) {
        for (int j = 1; j <= i; j++){
            System.out.print("*");
            }
        System.out.println("");
    }
}
public static void main(String[] args) {
    stars();
}
}

Answer 1

This is how I might write it.

// three loops
public static void stars(int size) {
    for (int y = 0; y < size; y++) {
        for (int i = 0; i < y && i < size - y - 1; i++)
            System.out.print(' ');
        for (int i = Math.min(y, size - y - 1); i < Math.max(y + 1, size - y); i++)
            System.out.print('*');
        System.out.println();
    }
}

or

// two loops
public static void stars(int size) {
    for (int y = 0; y < size; y++) {
        for (int x = 0; x < size; x++)
            System.out.print(
                    (x >= y && x < size - y) ||
                            (x >= size - y - 1 && x <= y) ? '*' : ' ');
        System.out.println();
    }
}

or

// one loop
public static void stars(int size) {
    for (int i = 0; i < size * size; i++) {
        int y = i / size, x = i % size;
        System.out.print(
                (x >= y && x < size - y) ||
                        (x >= size - y - 1 && x <= y) ? '*' : ' ');
        if (x == size - 1)
            System.out.println();
    }
}

Note: Whether this uses one, two or three loops, the time complexity is O(N^2). A simple way to determine this is the number of stars produced is O(N^2) no matter how it is done.