# 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 $O(1)$.

• 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;
}
}

• 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;
}
};

• 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


• 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 := range points {
i, j := p[0], p[1]
dx, dy := i-x, j-y
if dx*dx+dy*dy <= r*r {
cnt++
}
}
ans = append(ans, cnt)
}
return
}

• function countPoints(points: number[][], queries: number[][]): number[] {
return queries.map(([cx, cy, r]) => {
let res = 0;
for (const [px, py] of points) {
if (Math.sqrt((cx - px) ** 2 + (cy - py) ** 2) <= r) {
res++;
}
}
return res;
});
}


• impl Solution {
pub fn count_points(points: Vec<Vec<i32>>, queries: Vec<Vec<i32>>) -> Vec<i32> {
queries
.iter()
.map(|v| {
let cx = v[0];
let cy = v[1];
let r = v[2].pow(2);
let mut count = 0;
for p in points.iter() {
if (p[0] - cx).pow(2) + (p[1] - cy).pow(2) <= r {
count += 1;
}
}
count
})
.collect()
}
}