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Problem Statement

You are given an m x n binary matrix grid. We can flip the matrix’s rows and columns. A row or column flip means turning all the 0s into 1s and all the 1s into 0s. Our goal is to maximize the score after flipping.

The score of the matrix is calculated by treating each row as a binary number and interpreting it as an integer. The final score is the sum of these integers.

Clarifying Questions

  1. Can the grid be empty?
    • No, the problem statement ensures that the matrix has dimensions m x n where both m and n are at least 1.
  2. What should we prioritize, row flipping or column flipping?
    • The primary goal is to maximize the score. Generally, flipping rows first to maximize the most significant bits (i.e., leftmost bits) as 1 would be most beneficial. Following that, column flipping can help to ensure the maximum sum for the remaining bits.

Strategy

  1. Ensure the most significant bit (leftmost bit) of each row is 1:
    • Flip the entire row if the first element is 0.
  2. For the remaining column bits, optimize their count:
    • Count the number of 1s in each column (excluding the most significant bit).
    • If the count of 0s in a column is more than the count of 1s, flip that column.
  3. Final Calculation:
    • Compute the score of the matrix by interpreting each row as a binary number and summing these numbers.

Code

Let’s write the Python code to solve this:

def matrixScore(grid):
    # Step 1: Ensure the most significant bit (leftmost bit) is 1 for all rows
    for row in grid:
        if row[0] == 0:
            for j in range(len(row)):
                row[j] ^= 1
    
    # Step 2: Optimize the remaining columns (Excluding the first column)
    num_rows = len(grid)
    num_cols = len(grid[0])

    for j in range(1, num_cols):
        col_sum = sum(grid[i][j] for i in range(num_rows))
        # If more than half the rows have 0's in this column j, flip this column
        if col_sum < num_rows / 2:
            for i in range(num_rows):
                grid[i][j] ^= 1
    
    # Step 3: Calculate the final score
    score = 0
    for row in grid:
        # Interpret the binary number the row represents and add to the total score
        score += int(''.join(map(str, row)), 2)
    
    return score

Time Complexity

  1. Row Flipping: This takes O(m * n) as we may need to flip each bit in the row.
  2. Column Flipping: This also takes O(m * n) as we potentially flip each cell in the column.
  3. Score Calculation: This takes O(m * n) for constructing binary strings and converting them to integers.

Thus, the overall time complexity is O(m * n) where m is the number of rows and n is the number of columns. This is efficient given that each element in the grid needs to be considered for the optimal result.

This approach ensures we maximize the score by aiming to have the most significant bits as 1 and then optimizing the remaining bits to contribute maximally to the score.

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