Problem Description

This problem presents a geometric challenge involving both points on a plane and circles described by a center and a radius. You are provided with two arrays; the first, points, contains coordinates of various points on a 2D plane, and the second, queries, contains the specifications for several circles. Each entry in queries provides the central coordinates of a circle and its radius.

The task is to determine how many points from the points array fall within or on the boundary of each circle described in queries. To fall within or on the boundary, a point's distance from the center of the circle must be less than or equal to the circle's radius. For each circle in queries, the output should be the count of such points, and these counts are to be returned as an array.

The key aspect to consider here is the definition of a point being inside a circle. Geometrically, a point (x, y) is inside or on the boundary of a circle with center (xc, yc) and radius r if the distance from (x, y) to (xc, yc) is less than or equal to r. The distance between the two points is calculated using the Pythagorean theorem, which in this case does not require the square root calculation because we can compare the squares of the distances directly to the square of the radius.

Intuition

To find the intuitive solution to this problem, think about the standard way of measuring distance between two points on a plane - through the Pythagorean theorem. Usually, the formula √((x2 - x1)^2 + (y2 - y1)^2) is used, where (x1, y1) and (x2, y2) are coordinates of two points. In the context of a circle, a point lies inside or on the circle if this distance is less than or equal to the radius of the circle.

However, since comparing distances can be done without extracting the square root, the formula simplifies to (x2 - x1)^2 + (y2 - y1)^2 <= r^2. This squared comparison avoids unnecessary computation of square roots, makes the program run faster, and is helpful for counting points within the radius.

We arrive at our solution by iterating over each circle described in queries, and for each circle, we go through every point in the points array. For each point, we calculate this squared distance from the point to the circle's center and compare it to the square of the radius. If the squared distance is less than or equal to the squared radius, we increment a count. After checking all points for a particular circle, we append the total count to our answer array, and then proceed to the next circle.

The solution, therefore, follows a straight-forward brute-force approach. Its efficiency could, however, be improved by eliminating the need to calculate the distance for all points with respect to all circles, possibly by pre-sorting or partitioning the points.

Learn more about Math patterns.

Solution Approach

The solution provided in Python makes use of a direct implementation of the brute-force approach discussed in the intuition. It utilizes basic data structures from Python's standard library, namely lists, to store the sequence of points and queries and to accumulate the answer. The algorithm follows two nested loops to compare each point with every circle.

Here's a step by step explanation of the algorithm:

1. Initialize an empty list named ans which will store the number of points inside each circle.

   ans = []

2. The outer loop iterates over every circle query in the queries list. For each circle, the loop retrieves the center coordinates (x, y) and the radius r.

   for x, y, r in queries:

3. Inside the outer loop, we initialize a counter cnt to zero for counting how many points are within the current circle including on its boundary. This counter will be reset for each circle.

   cnt = 0

4. A nested inner loop runs through every point in points, where i and j are the x and y coordinates of the current point being checked.

   for i, j in points:

5. For each point, the solution calculates the squared distance from the point to the circle's center using the difference in x coordinates dx = i - x, the difference in y coordinates dy = j - y, and then summing their squares dx * dx + dy * dy.

   dx, dy = i - x, j - y

6. It then checks if this squared distance is less than or equal to the square of the radius r * r. If this condition is true, it means the point is inside or on the boundary of the circle, so it increments the counter cnt.

   cnt += dx * dx + dy * dy <= r * r

7. After the inner loop has finished checking all points, the inner loop ends, and the code appends the count for the current circle to the ans array.

   ans.append(cnt)

8. Finally, once all queries have been checked, the function returns the list ans which contains the count of points inside or on the boundary of each circle.

There are no additional data structures or sophisticated patterns employed in this solution. It relies on the fact that distance checking using squares avoids the need for math library calls, which improves computation time, but still, the solution has a time complexity of O(n * m), where n is the number of points and m is the number of queries, which is not efficient for larger datasets.

Example Walkthrough

Let's consider a simple example to illustrate the solution approach. Suppose we have the following points and queries:

• points = [(1, 3), (3, 3), (5, 3)]
• queries = [((2, 3), 2), ((4, 3), 1)]

We're supposed to find out how many points fall within or on the boundary of each circle described by the queries.

Query 1

• Center (x, y) = (2, 3), Radius (r) = 2

For each point, calculate the squared distance to the center and compare it to the squared radius (r^2 = 4):

1. Point (1, 3): dx = 1 - 2 = -1, dy = 3 - 3 = 0
   • Squared distance = (-1)^2 + 0^2 = 1
   • 1 <= 4 (True), so this point is inside the circle.
2. Point (3, 3): dx = 3 - 2 = 1, dy = 3 - 3 = 0
   • Squared distance = 1^2 + 0^2 = 1
   • 1 <= 4 (True), so this point is also inside the circle.
3. Point (5, 3): dx = 5 - 2 = 3, dy = 3 - 3 = 0
   • Squared distance = 3^2 + 0^2 = 9
   • 9 <= 4 (False), so this point is outside the circle.

There are 2 points inside or on the boundary of the first circle.

Query 2

• Center (x, y) = (4, 3), Radius (r) = 1

Perform the same steps with r^2 = 1:

1. Point (1, 3): dx = 1 - 4 = -3, dy = 3 - 3 = 0
   • Squared distance = (-3)^2 + 0^2 = 9
   • 9 <= 1 (False), so this point is outside the circle.
2. Point (3, 3): dx = 3 - 4 = -1, dy = 3 - 3 = 0
   • Squared distance = (-1)^2 + 0^2 = 1
   • 1 <= 1 (True), so this point is on the boundary of the circle.
3. Point (5, 3): dx = 5 - 4 = 1, dy = 3 - 3 = 0
   • Squared distance = 1^2 + 0^2 = 1
   • 1 <= 1 (True), so this point is also on the boundary of the circle.

There is 1 point inside or on the boundary of the second circle.

In summary, the counts for each query are [2, 1]. Thus, the final returned list of counts would be [2, 1], indicating that two points are within or on the boundary of the first described circle, and one point is within or on the boundary of the second described circle.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n * m), where n is the number of points and m is the number of queries. This is because there are two nested loops: the outer loop iterates over each query (which is m in number), and the inner loop iterates over each point (which is n in number). In the inner loop, we are doing a constant-time computation to check if a point is within the radius of the query circle.

Space Complexity

The space complexity of the code is O(m), where m is the number of queries. This is due to the fact that we are only using an additional list ans to store the results for each query. The size of this list grows linearly with the number of queries, therefore the space complexity is directly proportional to the number of queries.

Learn more about how to find time and space complexity quickly using problem constraints.