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1828. Queries on Number of Points Inside a Circle

Description

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)?

Solutions

Solution 1: Enumeration

Enumerate all the circles $(x, y, r)$. For each circle, calculate the number of points within the circle to get the answer.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of the arrays queries and points respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

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class Solution:
    def countPoints(
        self, points: List[List[int]], queries: List[List[int]]
    ) -> List[int]:
        ans = []
        for x, y, r in queries:
            cnt = 0
            for i, j in points:
                dx, dy = i - x, j - y
                cnt += dx * dx + dy * dy <= r * r
            ans.append(cnt)
        return ans

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class Solution {
    public int[] countPoints(int[][] points, int[][] queries) {
        int m = queries.length;
        int[] ans = new int[m];
        for (int k = 0; k < m; ++k) {
            int x = queries[k][0], y = queries[k][1], r = queries[k][2];
            for (var p : points) {
                int i = p[0], j = p[1];
                int dx = i - x, dy = j - y;
                if (dx * dx + dy * dy <= r * r) {
                    ++ans[k];
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
        vector<int> ans;
        for (auto& q : queries) {
            int x = q[0], y = q[1], r = q[2];
            int cnt = 0;
            for (auto& p : points) {
                int i = p[0], j = p[1];
                int dx = i - x, dy = j - y;
                cnt += dx * dx + dy * dy <= r * r;
            }
            ans.emplace_back(cnt);
        }
        return ans;
    }
};

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func countPoints(points [][]int, queries [][]int) (ans []int) {
    for _, q := range queries {
        x, y, r := q[0], q[1], q[2]
        cnt := 0
        for _, p :=