# LeetCode 1828. 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:**![](https://assets.leetcode.com/uploads/2021/03/25/chrome_2021-03-25_22-34-16.png)

```
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:**![](https://assets.leetcode.com/uploads/2021/03/25/chrome_2021-03-25_22-42-07.png)

```
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)`?

## Solution

```
class Solution {
public:
    vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
        vector<int> ans;
        
        for (const vector<int>& query : queries) {
            int num = 0;
            int threshold = query[2] * query[2];
            for (const vector<int>& point : points) {
                int dis_square = (point[0] - query[0]) * (point[0] - query[0]) + (point[1] - query[1]) * (point[1] - query[1]);
                if (dis_square <= threshold) {
                    num++;
                }
            }
            ans.push_back(num);
        }
        
        return ans;
    }
};
```
