1828. Queries on Number of Points Inside a Circle

QUESTION:

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 <= x​​​​​​i, y​​​​​​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)?

EXPLANATION:

思路也很简单, 就是两个for循环, 然后查看到圆心的距离是否小于半径.

SOLUTION:

class Solution {
    func countPoints(_ points: [[Int]], _ queries: [[Int]]) -> [Int] {
        var result:[Int] = []
        for query in queries {
            var count = 0
            for point in points {
                if ((query[0]-point[0])*(query[0]-point[0])+((query[1]-point[1])*(query[1]-point[1])) <= query[2]*query[2]) {
                    count += 1
                }
            }
            result.append(count)
        }
        return result
    }
}
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