Introduction

In this blog post, we will explore how to solve the LeetCode problem "1072" 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 matrix. You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa). Return the maximum number of rows that have all values equal after some number of flips. Example 1: Input: matrix = [[0,1],[1,1]] Output: 1 Explanation: After flipping no values, 1 row has all values equal. Example 2: Input: matrix = [[0,1],[1,0]] Output: 2 Explanation: After flipping values in the first column, both rows have equal values. Example 3: Input: matrix = [[0,0,0],[0,0,1],[1,1,0]] Output: 2 Explanation: After flipping values in the first two columns, the last two rows have equal values. Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 300 matrix[i][j] is either 0 or 1.

Explanation

Here's the solution:

• Key Idea: Two rows can become identical after flips if they are either the same or the bitwise inverse of each other.
• Hashing/Counting: Represent each row by its original form or its flipped form (flipping the first element). Count the occurrences of each of these forms.

• Maximization: The maximum count among all these forms is the answer.

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

Code

    def maxEqualRowsAfterFlips(matrix):
    """
    Finds the maximum number of rows that have all values equal after some number of flips.

    Args:
        matrix: A list of lists of integers representing the binary matrix.

    Returns:
        The maximum number of rows that have all values equal after some number of flips.
    """
    counts = {}
    for row in matrix:
        # Normalize the row: if the first element is 1, flip the row
        if row[0] == 1:
            normalized_row = tuple(1 - x for x in row)
        else:
            normalized_row = tuple(row)

        counts[normalized_row] = counts.get(normalized_row, 0) + 1

    return max(counts.values())