DEV Community

Abhishek Chaudhary
Abhishek Chaudhary

Posted on

Queries on Number of Points Inside a Circle

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 <= 500
  • points[i].length == 2
  • 0 <= xā€‹ā€‹ā€‹ā€‹ā€‹ā€‹i, yā€‹ā€‹ā€‹ā€‹ā€‹ā€‹i <= 500
  • 1 <= queries.length <= 500
  • queries[j].length == 3
  • 0 <= xj, yj <= 500
  • 1 <= rj <= 500
  • All coordinates are integers.

Follow up: Could you find the answer for each query in better complexity than O(n)?

SOLUTION:

class Solution:
    def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
        answer = []
        for x, y, r in queries:
            numPoints = 0
            for cx, cy in points:
                if (cx - x) ** 2 + (cy - y) ** 2 <= r ** 2:
                    numPoints += 1
            answer.append(numPoints)
        return answer
Enter fullscreen mode Exit fullscreen mode

Top comments (0)