861. Score After Flipping Matrix

861. Score After Flipping Matrix (Medium)

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.

Companies:
IIT Bombay

Related Topics:
Greedy

Solution 1.

1. Makes sure the first column only contains 1s by toggling a row if the first element in the row is 0.
2. For the remaining columns, if there are more 0s than 1s, toggle the column.

// OJ: https://leetcode.com/problems/score-after-flipping-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(1)
class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        for (int i = 0; i < M; ++i) {
            if (A[i][0]) continue;
            for (int j = 0; j < N; ++j) A[i][j] = 1 - A[i][j];
        }
        for (int j = 1; j < N; ++j) {
            int one = 0;
            for (int i = 0; i < M; ++i) one += A[i][j];
            if (one * 2 >= M) continue;
            for (int i = 0; i < M; ++i) A[i][j] = 1 - A[i][j];
        }
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) ans += A[i][j] * (1 << (N - j - 1));
        }
        return ans;
    }
};