Class Solution
java.lang.Object
g1801_1900.s1828_queries_on_number_of_points_inside_a_circle.Solution
1828 - Queries on Number of Points Inside a Circle\.
Medium
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 <= xi, yi <= 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)`?-
Constructor Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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countPoints
public int[] countPoints(int[][] points, int[][] queries)
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