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Leetcode 861. Score After Flipping Matrix

Problem Statement

You are given a matrix consisting of 0s and 1s, and you can perform any number of the following operations:

  1. Flip a row: Change all 0s to 1s and all 1s to 0s in that row.
  2. Flip a column: Change all 0s to 1s and all 1s to 0s in that column.

After each operation, every element of the matrix must still be either 0 or 1. You want to maximize the score after any number of operations. The score is obtained by interpreting each row of the matrix as a binary number, and then summing these numbers.

Your task is to return the maximum possible score.

Clarifying Questions

  1. Input Matrix Dimensions: Is there a guaranteed minimum or maximum for the dimensions of the input matrix?
    • Usually, this is not explicitly constrained, but typical constraints range for values that fit within reasonable computational limits.
  2. Matrix Contents: Is it always guaranteed that the matrix contains only 0s and 1s?
    • Yes, the problem states that the matrix consists only of 0s and 1s.
  3. Expected Output: How should the score be returned, integer, long, string…?
    • The score should be returned as an integer.
  4. Edge Cases: Should we consider edge cases such as the smallest matrix (1x1) or all elements being 0s?
    • Yes, you should consider edge cases in your implementation.

Strategy

To maximize the score:

  1. Ensure Left-most Column is Maximized: Convert the left-most bit of every row to 1 by flipping necessary rows. This ensures that each row’s value is at least half of the highest possible value.
  2. Maximize Column Values Starting from the Left: For each subsequent column, determine if flipping that column will result in more 1s than the original. If yes, perform the flip.

Code

Here’s the implementation of the strategy:

public class Solution {
    public int matrixScore(int[][] grid) {
        int rows = grid.length;
        int cols = grid[0].length;
        
        // Step 1: Ensure all rows have a leading 1 by flipping rows if necessary.
        for (int i = 0; i < rows; i++) {
            if (grid[i][0] == 0) {
                flipRow(grid, i);
            }
        }
        
        // Step 2: Optimize each column to maximize the number of 1s.
        for (int j = 1; j < cols; j++) {
            int numOnes = 0;
            // Count the number of 1s in the current column
            for (int i = 0; i < rows; i++) {
                if (grid[i][j] == 1) {
                    numOnes++;
                }
            }
            // If more than half of the column elements are 0s, flip the column
            if (numOnes < (rows + 1) / 2) {
                flipColumn(grid, j);
            }
        }
        
        // Step 3: Calculate the final score
        int score = 0;
        for (int i = 0; i < rows; i++) {
            int rowValue = 0;
            for (int j = 0; j < cols; j++) {
                rowValue = rowValue * 2 + grid[i][j];
            }
            score += rowValue;
        }
        
        return score;
    }
    
    private void flipRow(int[][] grid, int row) {
        for (int j = 0; j < grid[0].length; j++) {
            grid[row][j] ^= 1;
        }
    }
    
    private void flipColumn(int[][] grid, int col) {
        for (int i = 0; i < grid.length; i++) {
            grid[i][col] ^= 1;
        }
    }
}

Time Complexity

Overall, the time complexity is O(N * M), which is efficient for typical input sizes expected in competitive programming.

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