Java two-dimensional array filled with pattern

Question

how can I fill two-dimensional array of given "size" to get this output:
N=1

000
010
000

N=2

0000
0110
0110
0000

N=3

00000
01110
01210
01110
00000

public static void main (String[] args) throws java.lang.Exception
{
    int x, y;
    int N = 3;
    int counter = 0;
    x = y = N + 2;
    int[][] array = new int[x][y];
    for( int i = 0; i<x; i++){
        for( int j = 0; j < y; j++){
            if( i == 0 || i == x-1 || j == 0 || j == y-1){
                array[i][j] = 0;
            } else {
                array[i][j] = 1;
            }
            System.out.print(array[i][j]);
        }
        System.out.println();
    }
}

This code gives me just 1s surrounded by 0, but I cannot figure out how to increase value to the middle of array. If N is an odd number I can use modulo, but i don't know how to do this with even numbers.

Why the 3rd array has value 2 in the middle of array? What is the logic? — Emre Savcı
– Emre Savcı, Commented Aug 27, 2018 at 10:47
@EmreSavcı Smallest distance to the edge determines the number. Square size is 2 + N. This should answer the question too. — Torben
– Torben, Commented Aug 27, 2018 at 10:51
If I may suggest, your problem does not seem to come for your code, but from your algorithm. If you look closely you'll see that what you programmed will never what you expect on your display. — avi.elkharrat
– avi.elkharrat, Commented Aug 27, 2018 at 10:52

Sharon Ben Asher · Accepted Answer · 2018-08-27 10:52:09Z

3

The value of a cell is the shortest distance to the edge of the matrix. assuming cell[i,j], this can be either

i (distance from top)
j (distance from left)
N-i (distance from bottom)
N-j (distance from right)

so iterate over the matrix and calculate the shortest distance.

answered Aug 27, 2018 at 10:52

Sharon Ben Asher

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1 Comment

user2305208 Over a year ago

Thanks man, I didn't think of it as shortest distance from the edge, but it's the first thing you see when you look at this pattern.

tbsalling · Accepted Answer · 2018-08-27 11:22:45Z

Here's your homework :-)

import static java.lang.Math.min;

public static void main (String[] args) throws java.lang.Exception {
    int x, y;
    int N = 5;
    int counter = 0;
    x = y = N + 2;
    int[][] array = new int[x][y];
    for (int i = 0; i < x; i++) {
        counter = 0;
        for (int j = 0; j < y; j++) {
            if (i == 0 || i == x - 1 || j == 0 || j == y - 1) {
                array[i][j] = 0;
            } else {
                array[i][j] = min(i,min(min(i,N-i+1),min(j,N-j+1)));
            }
            System.out.print(array[i][j]);
        }
        System.out.println();
    }
}

It works by using

min(i,min(min(i,N-i+1),min(j,N-j+1)))

to calculate the shortest (minimum) distance to the edge. There are 4 quadrants in the matrix with each its own calculation of distance to edge - hence the four min-operations.

N=1:

000
010
000

N=2:

N=3:

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