Basically for each element in the upper-left quadrant, there is a possible of 4 values that can fit there (no flip, flip vertical, flip horizontal, flip both).

So just write a code to identify each group of 4 values, then calculate the max of each 4-value group, and sum everything up together.

The elements of each group should have the following indexes: - matrix[x][y], matrix[2n-1-x][y], matrix[x][2n-1-y], matrix[2n-1-x][2n-1-y] - where x & y are indexes between 0 and n-1

I just ran into this problem on a test that I had 24 min to solve. I didn't solve it in time. My brain spun to even see the pattern needed beyond getting the max value. So I went and learned the trick online... https://youtu.be/4rin1enhuQQ?si=By2NI2du9ZO6qtf7

https://github.com/SWT92/hackerrank_flipmatrix

My answer seems different from the originator, anyone got 483 the following matrix?

matrix = [[112, 42, 83, 119], [56, 125, 56, 49],[15, 78, 101, 43], [62, 98, 114, 108]]

this problem deep fried my brain

Symmetry: We exploit the symmetry of the 2Nx2N matrix to reduce calculations. Maximum value: For each position inside N x N, we find the maximum among the number and its three symmetric counterparts. Summation: The sum of these maximum values gives us the desired result.

    public static int flippingMatrix(List<List<Integer>> matrix) {
        int n = matrix.get(0).size()/2;
        int result = 0; 
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<n; j++){
                result += Math.max(matrix.get(i).get(j), Math.max(matrix.get(2*n-1-i).get(j),
                Math.max(matrix.get(2*n-1-i).get(2*n-1-j), matrix.get(i).get(2*n-1-j))));
            }
        }
        return result;
    }