Score After Flipping Matrix

LeetCode Q 861 - Score After Flipping Matrix

We have a two dimensional matrix A where each value is 0 or 1.
A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s.
After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.
Return the highest possible score.

Example 1:
Input: [[0,0,1,1],[1,0,1,0],[1,1,0,0]] ; Output: 39
Explanation:
Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Note:

• 1 <= A.length <= 20
• 1 <= A[0].length <= 20
• A[i][j] is 0 or 1.

Solution

• A[i][0] is worth 1 << (N - 1) points, more than the sum of (A[i][1] + .. + A[i][N-1]). We need to toggle all A[i][0] to 1, here I toggle all lines for A[i][0] = 0.
• A[i][j] is worth 1 << (N - 1 - j).
  For every col, I count the current number of 1s.
  After step 1, A[i][j] becomes 1 if A[i][j] == A[i][0].
  if M - cur > cur, we can toggle this column to get more 1s.
  max(M, M - cur) will be the maximum number of 1s that we can get.

Code:

public int matrixScore(int[][] A) {

	int R = A.length, C = A[0].length, res = (1 << (C - 1)) * R;

	for (int c = 1; c < C; c++) {
		int count = 0; // number of 1s in each column
		for (int r = 0;  r < R; r++) {
			count += (A[r][c] == A[r][0] ? 1 : 0)
		}

		res += Math.max(count, R - count) * (1 << (C - 1 - c));
	}

	return res;
}