Introduction

In this blog post, we will explore how to solve the LeetCode problem "3239" 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 row or column is considered palindromic if its values read the same forward and backward. You can flip any number of cells in grid from 0 to 1, or from 1 to 0. Return the minimum number of cells that need to be flipped to make either all rows palindromic or all columns palindromic. Example 1: Input: grid = [[1,0,0],[0,0,0],[0,0,1]] Output: 2 Explanation: Flipping the highlighted cells makes all the rows palindromic. Example 2: Input: grid = [[0,1],[0,1],[0,0]] Output: 1 Explanation: Flipping the highlighted cell makes all the columns palindromic. Example 3: Input: grid = [[1],[0]] Output: 0 Explanation: All rows are already palindromic. Constraints: m == grid.length n == grid[i].length 1 <= m n <= 2 105 0 <= grid[i][j] <= 1

Explanation

Here's a solution to the problem, focusing on efficiency and optimality:

• Key Idea: The core idea is to recognize that we can independently minimize the number of flips needed to make either all rows palindromic OR all columns palindromic. We can compute both and then take the minimum. The crux of the row/column palindromization is that we only need to consider half of each row/column because the other half is determined by the first half in a palindromic arrangement.

• Optimization: To avoid redundant computations for palindromizing rows and columns, we perform the palindrome checks and flip calculations in separate functions for rows and columns.

• Minimum Flips: The minimum number of flips to make a row/column palindromic is determined by comparing elements from both ends, moving inwards. If elements at symmetrical positions are different, a flip is required.

• Complexity:

  • Runtime: O(m * n)
  • Storage: O(1)

Code

    def min_flips_to_palindromic(grid):
    """
    Calculates the minimum number of cell flips needed to make either all rows or all columns palindromic.

    Args:
        grid: A 2D list of integers representing the binary matrix.

    Returns:
        The minimum number of flips needed.
    """

    def min_flips_rows(grid):
        """Calculates minimum flips to make all rows palindromic."""
        m = len(grid)
        n = len(grid[0])
        flips = 0
        for i in range(m):
            for j in range(n // 2):
                if grid[i][j] != grid[i][n - 1 - j]:
                    flips += 1
        return flips

    def min_flips_cols(grid):
        """Calculates minimum flips to make all columns palindromic."""
        m = len(grid)
        n = len(grid[0])
        flips = 0
        for j in range(n):
            for i in range(m // 2):
                if grid[i][j] != grid[m - 1 - i][j]:
                    flips += 1
        return flips

    return min(min_flips_rows(grid), min_flips_cols(grid))