Score After Flipping Matrix

class Solution {
public:
    int computeMatrixScore(vector<vector<int>>& matrix) {
        int rows = matrix.size();
        int cols = matrix[0].size();

        // Initialize score using the first column's potential max value
        int resultScore = (1 << (cols - 1)) * rows;

        // Evaluate each column starting from the second
        for (int col = 1; col < cols; col++) {
            int countMatchingFirstRow = 0;
            for (int row = 0; row < rows; row++) {
                // Compare current element to the first element of the row
                if (matrix[row][col] == matrix[row][0]) {
                    countMatchingFirstRow++;
                }
            }
            // Maximize the count for either matching or flipping
            countMatchingFirstRow = max(countMatchingFirstRow, rows - countMatchingFirstRow);
            // Calculate score for this column and accumulate
            int contribution = (1 << (cols - col - 1)) * countMatchingFirstRow;
            resultScore += contribution;
        }

        return resultScore;
    }
};

This C++ solution details an approach to maximize the score of a binary matrix by potentially flipping each row and evaluating each column for a scoring method. The objective is to ensure that the first column contains all ones for maximum possible score, given that flipping a row inverts all its bits.

The solution consists of the following:

• Begin by initializing the score with the maximum possible value from the first column. This establishment of an initial score hinges on converting all zeros in the first column to ones either by direct counting or through flipping.

• Proceed to evaluate each subsequent column to maximize the ones count. By comparing each element in a column to the first element of its respective row, determine how closely elements align with the first column. Contingent on these findings:

  • Calculate how many elements match the first column directly.
  • Determine whether flipping non-matching elements in the rest of the rows would yield a higher count of ones.
  • The larger count from the above assessments gets used to compute the score contributions from this particular column.

• For each column, convert the count of ones into a score addition using a left bitwise shift operation, where the shift magnitude diminishes by one with each next column from left to right. This shift operation effectively doubles the value of ones in more significant columns.

• Aggregate these contributions to the initial score for final tallying.

This approach ensures a thorough examination and calculation of the potential highest score by considering both direct and flipped states for each row in the matrix, optimizing on a column-by-column basis.

class Solution {

    public int calculateMatrixScore(int[][] matrix) {
        int rows = matrix.length;
        int cols = matrix[0].length;

        int totalScore = (1 << (cols - 1)) * rows;

        for (int col = 1; col < cols; col++) {
            int numMatchingFirstElement = 0;
            for (int row = 0; row < rows; row++) {
                if (matrix[row][col] == matrix[row][0]) {
                    numMatchingFirstElement++;
                }
            }
            
            numMatchingFirstElement = Math.max(numMatchingFirstElement, rows - numMatchingFirstElement);
            int scoreForColumn = (1 << (cols - col - 1)) * numMatchingFirstElement;
            totalScore += scoreForColumn;
        }

        return totalScore;
    }
}

The given Java solution calculates the maximum score after flipping a binary matrix. The goal is to flip rows and columns in a manner that maximizes the overall matrix score where the score is calculated based on the binary numbers represented by each row.

• The solution begins by determining the dimensions of the matrix with rows referring to the number of rows and cols referring to the number of columns.
• The initial total score is calculated assuming all rows are flipped to have their most significant bits (leftmost column) as 1. Each '1' in the most significant bit contributes significantly (exponential value) to the row's score. This is ensured by setting every MSB to '1' through row flips where necessary.
• Iterate over each column starting from the second one. For each column:
  • Count how many elements match the first column's value. This counts helps decide whether to flip this column.
  • Maximize the number of '1's in the column by comparing the count of matching elements to its complement. This is done using the expression Math.max(numMatchingFirstElement, rows - numMatchingFirstElement).
  • Calculate the score for that column based on the number of '1's adjusted post any potential flip, which is added to the total score.
• Finally, the function returns the totalScore, which represents the highest score achievable with optimal row and column flips based on the matrix's initial state.

This approach effectively uses bitwise operations and greedy strategies to optimize the scoring, leveraging the positional significance of bits within the rows.

class Solution:
    def calculateMaxMatrixScore(self, matrix: List[List[int]]) -> int:
        rows = len(matrix)
        cols = len(matrix[0])
        
        # Calculate the initial score based on the most significant bit
        total_score = (1 << (cols - 1)) * rows
        
        # Iterate through each column starting from the second
        for col in range(1, cols):
            count = 0
            for row in range(rows):
                # Count if current bit is the same as the most significant bit
                if matrix[row][col] == matrix[row][0]:
                    count += 1
            
            # Maximize the count with its complement
            max_count = max(count, rows - count)
            # Update score using the value of the current bit position
            total_score += (1 << (cols - col - 1)) * max_count
        
        return total_score

This Python solution is designed to maximize the score of a binary matrix after potential row flips. The score is calculated based on a binary representation of the rows, where each row contributes largely through its most significant bit. Here is a breakdown of how the code operates:

• Calculate initial dimensions of the matrix and derive the initial score solely from the most significant bit (MSB). This MSB, being either 0 or 1, is critical as flipping rows can toggle these bits to maximize the row's numeric value.

• Iterate over each column of the matrix, except the first one. For each column:

  • Count the number of bits corresponding to the MSB of their respective row. This is under the assumption that aligning all bits in a column to either all 0s or all 1s will maximize that column's contribution to the total score.
  • Compare this count to its complement (total number of rows minus the count) to determine which is greater. This step decides if flipping bits in that column will increase the overall score.
  • Update the total score by adding the value of this column calculated at the power of two position of the column index. This echoes the weight of bits in binary numbers where the rightmost bit has the least weight.

• At the end of iteration, the function returns the total maximum score which indicates the highest possible value of the matrix, taking potential flips into account.

This approach ensures that each column’s contribution to the final score is optimized, considering how a bit aligns with the MSB of each row after potential flips. By harnessing binary operations and focusing on the relative weight of bits from left to right, the solution efficiently calculates the maximum achievable score.