Welcome to Subscribe On Youtube

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()
        }
    }
    
    

All Problems

All Solutions