Introduction

In this blog post, we will explore how to solve the LeetCode problem "766" 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

Given an m x n matrix, return true if the matrix is Toeplitz. Otherwise, return false. A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements. Example 1: Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]] Output: true Explanation: In the above grid, the diagonals are: "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]". In each diagonal all elements are the same, so the answer is True. Example 2: Input: matrix = [[1,2],[2,2]] Output: false Explanation: The diagonal "[1, 2]" has different elements. Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 20 0 <= matrix[i][j] <= 99 Follow up: What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once? What if the matrix is so large that you can only load up a partial row into the memory at once?

Explanation

• Iterate through diagonals: Check each diagonal to see if all elements are the same. Start diagonals from the first column and first row.
  • Early exit: Return false immediately if any diagonal violates the Toeplitz condition.
  • Efficient comparison: Directly compare adjacent elements along each diagonal.

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

Code

    def isToeplitzMatrix(matrix):
    """
    Determines if a matrix is Toeplitz.

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

    Returns:
        True if the matrix is Toeplitz, False otherwise.
    """
    rows = len(matrix)
    cols = len(matrix[0])

    for i in range(rows):
        for j in range(cols):
            if i > 0 and j > 0 and matrix[i][j] != matrix[i-1][j-1]:
                return False
    return True