Introduction

In this blog post, we will explore how to solve the LeetCode problem "2249" 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 a 2D integer array circles where circles[i] = [xi, yi, ri] represents the center (xi, yi) and radius ri of the ith circle drawn on a grid, return the number of lattice points that are present inside at least one circle. Note: A lattice point is a point with integer coordinates. Points that lie on the circumference of a circle are also considered to be inside it. Example 1: Input: circles = [[2,2,1]] Output: 5 Explanation: The figure above shows the given circle. The lattice points present inside the circle are (1, 2), (2, 1), (2, 2), (2, 3), and (3, 2) and are shown in green. Other points such as (1, 1) and (1, 3), which are shown in red, are not considered inside the circle. Hence, the number of lattice points present inside at least one circle is 5. Example 2: Input: circles = [[2,2,2],[3,4,1]] Output: 16 Explanation: The figure above shows the given circles. There are exactly 16 lattice points which are present inside at least one circle. Some of them are (0, 2), (2, 0), (2, 4), (3, 2), and (4, 4). Constraints: 1 <= circles.length <= 200 circles[i].length == 3 1 <= xi, yi <= 100 1 <= ri <= min(xi, yi)

Explanation

Here's the breakdown of the solution:

• Iterate over possible lattice points: Determine a bounding box encompassing all circles. Iterate through each integer coordinate (lattice point) within this box.
• Check point inclusion in circles: For each lattice point, check if it falls inside any of the given circles.

• Count distinct points: Maintain a set to store the lattice points found within at least one circle, ensuring that each point is counted only once.

• Runtime Complexity: O(n m k), where n is the number of circles, and m is the area of the bounding box, and k is the number of circles. In the worst-case scenario, k could be close to n, resulting in a complexity of O(n^2 m). *Storage Complexity: O(m), where m is the number of lattice points inside the circles.

Code

    def countLatticePoints(circles):
    """
    Counts the number of lattice points inside at least one circle.

    Args:
        circles: A list of circles, where each circle is represented as [x, y, r].

    Returns:
        The number of lattice points inside at least one circle.
    """

    points = set()
    for x_c, y_c, r in circles:
        for x in range(x_c - r, x_c + r + 1):
            for y in range(y_c - r, y_c + r + 1):
                if (x - x_c)**2 + (y - y_c)**2 <= r**2:
                    points.add((x, y))
    return len(points)