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. 1 <= A.length <= 20
2. 1 <= A[0].length <= 20
3. A[i][j] is 0 or 1.

 

Approach #1: C++.

class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int r = A.size(), c = A[0].size();
        int ans = 0;
        for (int i = 0; i < c; ++i) {
            int col = 0;
            for (int j = 0; j < r; ++j)
                col += A[j][i] ^ A[j][0];       //相同为0， 不同为1。
            ans += max(col, r - col) * (1 << (c - 1 - i));
        }
        return ans;
    }
};

　　

Analysis:

Because every row of the matrix repersent a binary number, we want to the number become bigger, so the left-most column we can make it become 1. The others columns we can make it become 1 as more as possible by flipping the column. we count the number of how many number in the column is same as the left-most column. and using max(col, r - col) to get the number of 1 we can get.