Introduction

In this blog post, we will explore how to solve the LeetCode problem "3122" 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 a 2D matrix grid of size m x n. In one operation, you can change the value of any cell to any non-negative number. You need to perform some operations such that each cell grid[i][j] is: Equal to the cell below it, i.e. grid[i][j] == grid[i + 1][j] (if it exists). Different from the cell to its right, i.e. grid[i][j] != grid[i][j + 1] (if it exists). Return the minimum number of operations needed. Example 1: Input: grid = [[1,0,2],[1,0,2]] Output: 0 Explanation: All the cells in the matrix already satisfy the properties. Example 2: Input: grid = [[1,1,1],[0,0,0]] Output: 3 Explanation: The matrix becomes [[1,0,1],[1,0,1]] which satisfies the properties, by doing these 3 operations: Change grid[1][0] to 1. Change grid[0][1] to 0. Change grid[1][2] to 1. Example 3: Input: grid = [[1],[2],[3]] Output: 2 Explanation: There is a single column. We can change the value to 1 in each cell using 2 operations. Constraints: 1 <= n, m <= 1000 0 <= grid[i][j] <= 9

Explanation

Here's the breakdown:

• Enforce Column Consistency: Iterate through the matrix, ensuring that each column has identical values.
• Enforce Row Difference: Then, iterate through each row, making sure adjacent cells have different values. If they don't, change the right cell to a value different from its left neighbor.

• Count Operations: Keep track of the number of changes (operations) made.

• Runtime Complexity: O(m n), where m is the number of rows and n is the number of columns. *Storage Complexity: O(1)

Code

    def min_operations(grid):
    m = len(grid)
    n = len(grid[0])
    operations = 0

    # Enforce column consistency
    for j in range(n):
        for i in range(1, m):
            if grid[i][j] != grid[i-1][j]:
                grid[i][j] = grid[i-1][j]
                operations += 1

    # Enforce row difference
    for i in range(m):
        for j in range(n - 1):
            if grid[i][j] == grid[i][j+1]:
                # Change grid[i][j+1] to a different value
                for k in range(10):
                    if k != grid[i][j]:
                        grid[i][j+1] = k
                        operations += 1
                        break

    return operations