Introduction

In this blog post, we will explore how to solve the LeetCode problem "3000" 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 0-indexed integer array dimensions. For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i. Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area. Example 1: Input: dimensions = [[9,3],[8,6]] Output: 48 Explanation: For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 9 + 3 3) = sqrt(90) ≈ 9.487. For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 8 + 6 6) = sqrt(100) = 10. So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48. Example 2: Input: dimensions = [[3,4],[4,3]] Output: 12 Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12. Constraints: 1 <= dimensions.length <= 100 dimensions[i].length == 2 1 <= dimensions[i][0], dimensions[i][1] <= 100

Explanation

Here's the breakdown of the solution:

• Calculate Diagonal Squared: To avoid the computationally expensive sqrt operation, we'll compare the squares of the diagonals. The rectangle with the largest diagonal squared will have the largest diagonal.
• Track Largest Diagonal and Area: Iterate through the dimensions array, calculating the diagonal squared for each rectangle. Keep track of the largest diagonal squared encountered so far and the corresponding area.

• Handle Ties: If we find a rectangle with the same diagonal squared as the largest seen so far, we compare their areas and update the largest area if necessary.

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

Code

    def area_of_max_diagonal(dimensions):
    """
    Calculates the area of the rectangle with the longest diagonal.

    Args:
        dimensions: A 2D list of integers representing the length and width of rectangles.

    Returns:
        The area of the rectangle with the longest diagonal.
    """

    max_diagonal_squared = -1
    max_area = -1

    for length, width in dimensions:
        diagonal_squared = length * length + width * width
        area = length * width

        if diagonal_squared > max_diagonal_squared:
            max_diagonal_squared = diagonal_squared
            max_area = area
        elif diagonal_squared == max_diagonal_squared:
            max_area = max(max_area, area)

    return max_area