LC 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.

Runtime: 4 ms, faster than 58.17% of C++ online submissions for Score After Flipping Matrix.

注意一下，首列如果某行从0换成1那么在之后计算这一行时要反向。

class Solution {
public:
  int matrixScore(vector<vector<int>>& A) {
    int r = A.size();
    int c = A[0].size();
    vector<int> rsign(r,0);
    int base = 1 << (A[0].size()-1), ret = 0;
    for(int j=0; j<c; j++){
      int cntzero = 0, cntone = 0;
      for(int i=0; i<r; i++){
        if(j == 0){
          if(A[i][j] != 0) rsign[i]++;
        }else{
          if((A[i][j] == 0 && rsign[i] == 0) || (A[i][j] == 1 && rsign[i] == 1)) cntzero++;
          else cntone++;
        }
      }
      if(j == 0){
        ret += base * r;
      } else if(cntzero > cntone) {
        ret += base * cntzero;
      }else ret += base * cntone;
      base >>= 1;
    }
    
    return ret;
  }
};