861. Score After Flipping Matrix

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

You are given an m x n binary matrix grid.

A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0's to 1's, and all 1's to 0's).

Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score after making any number of moves (including zero moves).

Example 1:

Input: grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39
Explanation: 0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Example 2:

Input: grid = [[0]]
Output: 1

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 20
grid[i][j] is either 0 or 1.

🧠 Thuật Toán & Kỹ Thuật

Bit Manipulation (Thao tác bit)Matrix (Ma trận)

⏱️ Thời gian O(n²)

💾 Không gian O(n)

💻 Lời Giải

C++ 0861-score-after-flipping-matrix.cpp

class Solution {
public:
    int matrixScore(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        
        for (int i = 0; i < n; ++i) {
            if (grid[i][0] == 0) {
                for (int j = 0; j < m; ++j) {
                    grid[i][j] ^= 1;
                }
            }
        }
        
        for (int j = 0; j < m; ++j) {
            int cntOne = 0;
            for (int i = 0; i < n; ++i) {
                cntOne += grid[i][j];
            }
            if (cntOne < n - cntOne) {
                for (int i = 0; i < n; ++i) {
                    grid[i][j] ^= 1;
                }
            }
        }
        
        int ans = 0;
        
        for (int i = 0; i < n; ++i) {
            int dec = 0, mul = 1;
            for (int j = m - 1; j >= 0; --j) {
                dec += grid[i][j] * mul;
                mul *= 2;
            }
            ans += dec;
        }
        
        return ans;
    }
};