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Formatted question description: https://leetcode.ca/all/861.html

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/
// 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;
    }
};

  • class Solution {
    public int matrixScore(int[][] A) {
        int rows = A.length, columns = A[0].length;
        for (int i = 0; i < rows; i++) {
            if (A[i][0] == 0) {
                for (int j = 0; j < columns; j++)
                    A[i][j] = 1 - A[i][j];
            }
        }
        for (int i = 1; i < columns; i++) {
            int ones = 0;
            for (int j = 0; j < rows; j++)
                ones += A[j][i];
            if (ones < rows - ones) {
                for (int j = 0; j < rows; j++)
                    A[j][i] = 1 - A[j][i];
            }
        }
        int score = 0;
        for (int i = 0; i < rows; i++) {
            int[] row = A[i];
            int curScore = 0;
            for (int j = 0; j < columns; j++) {
                curScore *= 2;
                curScore += row[j];
            }
            score += curScore;
        }
        return score;
    }
}

############

class Solution {
    public int matrixScore(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; ++i) {
            if (grid[i][0] == 0) {
                for (int j = 0; j < n; ++j) {
                    grid[i][j] ^= 1;
                }
            }
        }
        int ans = 0;
        for (int j = 0; j < n; ++j) {
            int cnt = 0;
            for (int i = 0; i < m; ++i) {
                cnt += grid[i][j];
            }
            ans += Math.max(cnt, m - cnt) * (1 << (n - j - 1));
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/score-after-flipping-matrix/
// 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;
    }
};

  • class Solution:
    def matrixScore(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        for i in range(m):
            if grid[i][0] == 0:
                for j in range(n):
                    grid[i][j] ^= 1
        ans = 0
        for j in range(n):
            cnt = sum(grid[i][j] for i in range(m))
            ans += max(cnt, m - cnt) * (1 << (n - j - 1))
        return ans

############

class Solution(object):
    def matrixScore(self, A):
        """
        :type A: List[List[int]]
        :rtype: int
        """
        rows, cols = len(A), len(A[0])
        for row in range(rows):
            if A[row][0] == 0:
                A[row] = self.toggle_row(A[row])
        for col in range(1, cols):
            col_array = [A[row][col] for row in range(rows)]
            sum_col_array = sum(col_array)
            if sum_col_array <= rows / 2:
                col_array = self.toggle_col(col_array)
            for row in range(rows):
                A[row][col] = col_array[row]
        bin_row = []
        for row in range(rows):
            bin_row.append(int("".join(map(str, A[row])), 2))
        return sum(bin_row)
                
    def toggle_row(self, row_array):
        return [0 if x == 1 else 1 for x in row_array]
    
    def toggle_col(self, col_array):
        return [0 if x == 1 else 1 for x in col_array]

  • func matrixScore(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	for i := 0; i < m; i++ {
		if grid[i][0] == 0 {
			for j := 0; j < n; j++ {
				grid[i][j] ^= 1
			}
		}
	}
	ans := 0
	for j := 0; j < n; j++ {
		cnt := 0
		for i := 0; i < m; i++ {
			cnt += grid[i][j]
		}
		if cnt < m-cnt {
			cnt = m - cnt
		}
		ans += cnt * (1 << (n - j - 1))
	}
	return ans
}

  • function matrixScore(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    for (let i = 0; i < m; ++i) {
        if (grid[i][0] == 0) {
            for (let j = 0; j < n; ++j) {
                grid[i][j] ^= 1;
            }
        }
    }
    let ans = 0;
    for (let j = 0; j < n; ++j) {
        let cnt = 0;
        for (let i = 0; i < m; ++i) {
            cnt += grid[i][j];
        }
        ans += Math.max(cnt, m - cnt) * (1 << (n - j - 1));
    }
    return ans;
}

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