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Formatted question description: https://leetcode.ca/all/1828.html

1828. Queries on Number of Points Inside a Circle

Level

Medium

Description

You are given an array points where points[i] = [x_i, y_i] is the coordinates of the i-th point on a 2D plane. Multiple points can have the same coordinates.

You are also given an array queries where queries[j] = [x_j, y_j, r_j] describes a circle centered at (x_j, y_j) with a radius of r_j.

For each query queries[j], compute the number of points inside the j-th circle. Points on the border of the circle are considered inside.

Return an array answer, where answer[j] is the answer to the j-th query.

Example 1:

Image text

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:

Image text

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 <= x_j, y_j <= 500
  • 1 <= r_j <= 500
  • All coordinates are integers.

Solution

For each query, obtain the values x, y and r, and for each point, calculate the squared distance between the query’s center and the point. If the squared distance is less than or equal to r * r, then the point is inside the circle. The number of points inside each circle can be calculated.

  • class Solution {
    public int[] countPoints(int[][] points, int[][] queries) {
        int queriesCount = queries.length;
        int[] counts = new int[queriesCount];
        for (int i = 0; i < queriesCount; i++) {
            int[] query = queries[i];
            int x = query[0], y = query[1], r = query[2];
            int rSquare = r * r;
            for (int[] point : points) {
                int distanceSquare = getDistanceSquare(point, x, y);
                if (distanceSquare <= rSquare)
                    counts[i]++;
            }
        }
        return counts;
    }

    public int getDistanceSquare(int[] point, int x, int y) {
        return (x - point[0]) * (x - point[0]) + (y - point[1]) * (y - point[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;
    }
}

  • // OJ: https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/
// Time: O(PQ)
// Space: O(1)
class Solution {
public:
    vector<int> countPoints(vector<vector<int>>& P, vector<vector<int>>& Q) {
        vector<int> ans;
        for (auto &q : Q) {
            int cnt = 0;
            for (auto &p : P) {
                if (pow(q[0] - p[0], 2) + pow(q[1] - p[1], 2) <= pow(q[2], 2)) ++cnt;
            }
            ans.push_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

############

# 1828. Queries on Number of Points Inside a Circle
# https://leetcode.com/problems/queries-on-number-of-points-inside-a-circle/

class Solution:
    def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
        res = []
        
        for x, y, r in queries:
            count = 0
            
            for p1,p2 in points:
                dx = abs(p1 - x)
                dy = abs(p2 - y)
                
                count += int(dx**2 + dy**2 <= r**2)
            
            res.append(count)
        
        return res


  • 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()
    }
}

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