You are given an array points where points[i] = [xi, yi] is the coordinates of the ith point on a 2D plane. Multiple points can have the same coordinates.
You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at (xj, yj) with a radius of rj.
For each query queries[j], compute the number of points inside the jth circle. Points on the border of the circle are considered inside.
Return an array answer, where answer[j] is the answer to the jth query.
Example 1:
Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]] Output: [3,2,2] Explanation: The points and circles are shown above. queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.
Example 2:
Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]] Output: [2,3,2,4] Explanation: The points and circles are shown above. queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.
Constraints:
1 <= points.length <= 500points[i].length == 20 <= xi, yi <= 5001 <= queries.length <= 500queries[j].length == 30 <= xj, yj <= 5001 <= rj <= 500Follow up: Could you find the answer for each query in better complexity than O(n)?
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force method for this problem involves checking each point against every circle individually. We see if each point falls within the boundaries of each circle and count the ones that do. Finally, we record these counts as the answer to the queries.
Here's how the algorithm would work step-by-step:
def count_points_inside_circles_brute_force(points, circles):
results = []
# Iterate through each circle
for circle_index in range(len(circles)):
circle_center_x = circles[circle_index][0]
circle_center_y = circles[circle_index][1]
circle_radius = circles[circle_index][2]
points_inside_count = 0
# Check each point against the current circle
for point_index in range(len(points)):
point_x = points[point_index][0]
point_y = points[point_index][1]
# Calculate the distance between point and circle center
distance_x = point_x - circle_center_x
distance_y = point_y - circle_center_y
distance = (distance_x**2 + distance_y**2)**0.5
# Check if the point is inside the circle
if distance <= circle_radius:
# Increment counter if point is within
points_inside_count += 1
# Store the result for the current circle
results.append(points_inside_count)
return resultsWe're trying to efficiently count how many points are inside each circle. The smart way to do this is to check each point against each circle one by one, but in an organized way that lets us avoid unnecessary calculations. We can reduce redundant calculations by checking only points inside a bounding square of the circle and using the distance formula.
Here's how the algorithm would work step-by-step:
def count_points(points, queries):
result = []
for circle_index in range(len(queries)):
circle_center_x, circle_center_y, circle_radius = queries[circle_index]
point_count = 0
# Define the bounds of the square containing the circle
min_x = circle_center_x - circle_radius
max_x = circle_center_x + circle_radius
min_y = circle_center_y - circle_radius
max_y = circle_center_y + circle_radius
for point_index in range(len(points)):
point_x, point_y = points[point_index]
# Only check points within the bounding square
if min_x <= point_x <= max_x and min_y <= point_y <= max_y:
# Calculate distance between point and circle center
distance = ((point_x - circle_center_x)**2 + (point_y - circle_center_y)**2)**0.5
# Check if the point is inside the circle
if distance <= circle_radius:
point_count += 1
result.append(point_count)
# Return the count of points inside each circle
return result| Case | How to Handle |
|---|---|
| Empty points array | Return an empty list of counts immediately as there are no points to check against any circles. |
| Empty queries array | Return an empty list of counts immediately as there are no circles to query. |
| Points with very large or very small coordinates (potential overflow) | Use long data type for distance calculation to prevent integer overflow, or use appropriate libraries to deal with arbitrary sized numbers. |
| Queries with very large or very small radius (potential overflow) | Use long data type for distance calculation to prevent integer overflow, or use appropriate libraries to deal with arbitrary sized numbers. |
| Points clustered far away from any circles | The solution correctly handles this by returning zero for each query as no points are inside the circles. |
| Points concentrated near the circle's boundary; floating point precision issues | Implement a small tolerance value (epsilon) when comparing the distance to the radius to account for floating-point precision errors. |
| Large number of points and queries; performance concerns | Optimize by considering spatial partitioning techniques (e.g., KD-trees) if the naive O(m*n) complexity is too slow. |
| Points with coordinates of zero | The distance calculation handles zero coordinates correctly by not causing divide by zero or other mathematical exceptions. |