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Have no idea what's your question but here is a short method that's placing a boarder to a matrix. You could use the a as index for the B array.

private static void placeBoarders(int[][] matrix, int n) {
    int a = 0;

    for (int i = 0; i < 4 * (n-1); i++) {
        switch (i/(n - 1)) {
            case 0:
                matrix[i%(n-1)][0] = a++;
                break;
            case 1:
                matrix[n-1][i%(n-1)] = a++;
                break;
            case 2:
                matrix[(n - 1) - i%(n-1)][n-1] = a++;
                break;
            case 3:
                matrix[0][(n-1) - i%(n-1)] = a++;
                break;
            default:
                throw new IndexOutOfBoundsException();
        }
    }
}

The logic I've used here is that if I split the boarder to four equal chunks and I start from (top left -> bottom left -> bottom right -> top right -> top left) then I will have four pieces of size n-1. If you write it dawn on a paper then it will take like 5 mins to calculate the indexes.

For the opposite direction (top left -> top right -> bottom right -> bottom left -> top left) you should swap the matrix indexes just like this:

matrix[i%(n-1)][0] >> matrix[0][i%(n-1)]
matrix[n-1][i%(n-1)] >> matrix[i%(n-1)][n-1]
matrix[(n - 1) - i%(n-1)][n-1] >> matrix[n-1][(n - 1) - i%(n-1)]
matrix[0][(n-1) - i%(n-1)] >> matrix[(n-1) - i%(n-1)][0]