Introduction

In this blog post, we will explore how to solve the LeetCode problem "498" 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 mat, return an array of all the elements of the array in a diagonal order. Example 1: Input: mat = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,4,7,5,3,6,8,9] Example 2: Input: mat = [[1,2],[3,4]] Output: [1,2,3,4] Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 104 1 <= m * n <= 104 -105 <= mat[i][j] <= 105

Explanation

Here's a breakdown of the solution and the Python code:

• High-Level Approach:

  • Iterate through the diagonals. The number of diagonals is m + n - 1.
  • For each diagonal, determine the starting coordinates (row, col). The sum of the row and column indices for a given diagonal is constant.
  • Traverse the elements along the diagonal, adding them to the result. Reverse the order of elements in even-numbered diagonals to achieve the desired zig-zag pattern.

• Complexity:

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

Code

    def findDiagonalOrder(mat: list[list[int]]) -> list[int]:
    """
    Given an m x n matrix mat, return an array of all the elements of the array in a diagonal order.
    """
    if not mat or not mat[0]:
        return []

    m, n = len(mat), len(mat[0])
    result = []

    for d in range(m + n - 1):
        # Determine the starting coordinates for the diagonal
        if d < n:
            row, col = 0, d
        else:
            row, col = d - n + 1, n - 1

        diagonal_elements = []
        while row < m and col >= 0:
            diagonal_elements.append(mat[row][col])
            row += 1
            col -= 1

        # Reverse the order of elements for even-numbered diagonals
        if d % 2 == 0:
            diagonal_elements.reverse()

        result.extend(diagonal_elements)

    return result