Introduction

In this blog post, we will explore how to solve the LeetCode problem "861" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the approach, complexity analysis, and Python code for solving the binary matrix score maximization problem:

• Maximize Most Significant Bit: The most significant bit of each row contributes the most to the score. Therefore, ensure the first column has all 1s. Toggle rows as necessary to achieve this.
• Maximize Remaining Columns: For each remaining column, count the number of 1s. If the count is less than half the number of rows, toggle the column to get more 1s.

• Calculate Score: Convert each row to its integer representation and sum these values.

• Runtime Complexity: O(m*n) , where m is the number of rows and n is the number of columns.

• Storage Complexity: O(1)

Code

    class Solution:
    def matrixScore(self, grid: list[list[int]]) -> int:
        m = len(grid)
        n = len(grid[0])

        # Ensure the first column has all 1s
        for i in range(m):
            if grid[i][0] == 0:
                for j in range(n):
                    grid[i][j] ^= 1  # Toggle the row

        # Maximize remaining columns
        for j in range(1, n):
            ones = 0
            for i in range(m):
                ones += grid[i][j]
            if ones < m / 2:
                for i in range(m):
                    grid[i][j] ^= 1  # Toggle the column

        # Calculate the score
        score = 0
        for i in range(m):
            row_value = 0
            for j in range(n):
                row_value = (row_value << 1) | grid[i][j]
            score += row_value

        return score