Introduction

In this blog post, we will explore how to solve the LeetCode problem "3240" 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 all rows and columns palindromic, and the total number of 1's in grid divisible by 4. Example 1: Input: grid = [[1,0,0],[0,1,0],[0,0,1]] Output: 3 Explanation: Example 2: Input: grid = [[0,1],[0,1],[0,0]] Output: 2 Explanation: Example 3: Input: grid = [[1],[1]] Output: 2 Explanation: Constraints: m == grid.length n == grid[i].length 1 <= m n <= 2 105 0 <= grid[i][j] <= 1

Explanation

Here's a breakdown of the solution strategy:

• Palindrome Conversion: Focus on converting rows and columns into palindromes independently. The minimum flips for a row/column to become a palindrome is determined by comparing elements from the start and end, moving inwards.
• Parity Adjustment: Calculate the total number of 1s in the grid. If the count isn't divisible by 4, determine the minimum flips needed to make it so (either adding or removing 1s). Exploit the fact we can flip bits we already flipped for palindromes.

• Optimization: Iterate through each row and column to determine minimum flips needed for palindromic transformation. Keep track of the total number of ones and increment/decrement to achieve divisibility by 4, utilizing flips made during palindrome creation.

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

Code

    def min_flips(grid):
    m = len(grid)
    n = len(grid[0])
    row_flips = 0
    col_flips = 0
    total_ones = 0

    # Calculate row flips
    for i in range(m):
        for j in range(n // 2):
            if grid[i][j] != grid[i][n - 1 - j]:
                row_flips += 1

    # Calculate column flips
    for j in range(n):
        for i in range(m // 2):
            if grid[i][j] != grid[m - 1 - i][j]:
                col_flips += 1

    # Count total ones
    for i in range(m):
        for j in range(n):
            total_ones += grid[i][j]

    total_flips = row_flips + col_flips

    # Adjust for divisibility by 4
    remainder = total_ones % 4
    if remainder != 0:
        flips_needed = min(remainder, 4 - remainder)
        return total_flips + flips_needed

    return total_flips