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 <= xi, yi <= 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 $O(1)$.
/** * Note: The returned array must be malloced, assume caller calls free(). */int*countPoints(int**points,intpointsSize,int*pointsColSize,int**queries,intqueriesSize,int*queriesColSize,int*returnSize){int*ans=malloc(sizeof(int)*queriesSize);for(inti=0;i<queriesSize;i++){intcx=queries[i][0];intcy=queries[i][1];intr=queries[i][2];intcount=0;for(intj=0;j<pointsSize;j++){if(sqrt(pow(points[j][0]-cx,2)+pow(points[j][1]-cy,2))<=r){count++;}}ans[i]=count;}*returnSize=queriesSize;returnans;}