Traverse Matrix in Diagonal strips

Question

I thought this problem had a trivial solution, couple of for loops and some fancy counters, but apparently it is rather more complicated.

So my question is, how woul

Answer 1

Let's take a look how matrix elements are indexed.

(0,0)   (0,1)   (0,2)   (0,3)   (0,4)  
(1,0)   (1,1)   (1,2)   (1,3)   (1,4)  
(2,0)   (2,1)   (2,2)   (2,3)   (2,4)  

Now, let's take a look at the stripes:

Stripe 1: (0,0)
Stripe 2: (0,1)    (1,0)  
Stripe 3: (0,2)    (1,1)    (2,0)
Stripe 4: (0,3)    (1,2)    (2,1)
Stripe 5: (0,4)    (1,3)    (2,2)
Stripe 6: (1,4)    (2,3)
Stripe 7: (2,4)

If you take a closer look, you'll notice one thing. The sum of indexes of each matrix element in each stripe is constant. So, here's the code that does this.

public static void printSecondaryDiagonalOrder(int[][] matrix) {
    int rows = matrix.length;
    int cols = matrix[0].length;
    int maxSum = rows + cols - 2;

    for (int sum = 0; sum <= maxSum; sum++) {
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (i + j - sum == 0) {
                    System.out.print(matrix[i][j] + "\t");
                }
            }
        }
        System.out.println();
    }
}

It's not the fastest algorithm out there (does(rows * cols * (rows+cols-2)) operations), but the logic behind it is quite simple.