# Solving Leetcode Interviews in Seconds with AI: Rectangle Area


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "223" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 the coordinates of two rectilinear rectangles in a 2D plane, return the total area covered by the two rectangles. The first rectangle is defined by its bottom-left corner (ax1, ay1) and its top-right corner (ax2, ay2). The second rectangle is defined by its bottom-left corner (bx1, by1) and its top-right corner (bx2, by2).   Example 1:   Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2 Output: 45  Example 2:  Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2 Output: 16    Constraints:  -104 <= ax1 <= ax2 <= 104 -104 <= ay1 <= ay2 <= 104 -104 <= bx1 <= bx2 <= 104 -104 <= by1 <= by2 <= 104  

	# Explanation
	Here's the breakdown of the solution:

*   **Calculate Individual Areas:** Calculate the areas of the two rectangles separately using the formula: `area = (x2 - x1) * (y2 - y1)`.
*   **Calculate Intersection Area:** Determine the area of the overlapping region (if any) between the two rectangles. The intersection is itself a rectangle. Find the coordinates of the bottom-left and top-right corners of this intersection rectangle. If there's no overlap, the intersection area is zero.
*   **Calculate Total Area:** The total area is the sum of the individual areas minus the intersection area to avoid double-counting the overlapping region.

*   **Runtime Complexity:** O(1), **Storage Complexity:** O(1)

	
	# Code
	```python
	def compute_area(ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
    """
    Calculates the total area covered by two rectilinear rectangles in a 2D plane.

    Args:
        ax1: The x-coordinate of the bottom-left corner of the first rectangle.
        ay1: The y-coordinate of the bottom-left corner of the first rectangle.
        ax2: The x-coordinate of the top-right corner of the first rectangle.
        ay2: The y-coordinate of the top-right corner of the first rectangle.
        bx1: The x-coordinate of the bottom-left corner of the second rectangle.
        by1: The y-coordinate of the bottom-left corner of the second rectangle.
        bx2: The x-coordinate of the top-right corner of the second rectangle.
        by2: The y-coordinate of the top-right corner of the second rectangle.

    Returns:
        The total area covered by the two rectangles.
    """

    area_a = (ax2 - ax1) * (ay2 - ay1)
    area_b = (bx2 - bx1) * (by2 - by1)

    # Calculate intersection area
    left = max(ax1, bx1)
    right = min(ax2, bx2)
    bottom = max(ay1, by1)
    top = min(ay2, by2)

    intersection_width = max(0, right - left)
    intersection_height = max(0, top - bottom)
    intersection_area = intersection_width * intersection_height

    total_area = area_a + area_b - intersection_area

    return total_area
	```