Introduction

In this blog post, we will explore how to solve the LeetCode problem "3142" 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. You need to check if 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 true if all the cells satisfy these conditions, otherwise, return false. Example 1: Input: grid = [[1,0,2],[1,0,2]] Output: true Explanation: All the cells in the grid satisfy the conditions. Example 2: Input: grid = [[1,1,1],[0,0,0]] Output: false Explanation: All cells in the first row are equal. Example 3: Input: grid = [[1],[2],[3]] Output: false Explanation: Cells in the first column have different values. Constraints: 1 <= n, m <= 10 0 <= grid[i][j] <= 9

Explanation

Here's the breakdown of the solution:

• Iterate through the grid: The code will iterate through each cell of the given grid to check the specified conditions.
• Check conditions: For each cell, it verifies if the value is equal to the cell below it (if it exists) and different from the cell to its right (if it exists).

• Early exit: If any of the conditions are not met, the function immediately returns False. If the entire grid is traversed without finding any violations, it returns True.

• 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 check_grid_conditions(grid):
    """
    Checks if each cell in the grid satisfies the given conditions.

    Args:
        grid: A 2D matrix (list of lists) representing the grid.

    Returns:
        True if all cells satisfy the conditions, False otherwise.
    """
    m = len(grid)
    n = len(grid[0]) if m > 0 else 0  # Handle empty grid case

    for i in range(m):
        for j in range(n):
            # Check condition 1: Equal to the cell below (if exists)
            if i + 1 < m:
                if grid[i][j] != grid[i + 1][j]:
                    return False

            # Check condition 2: Different from the cell to the right (if exists)
            if j + 1 < n:
                if grid[i][j] == grid[i][j + 1]:
                    return False

    return True