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.
Solution: Brute Force
Time complexity: O(P * Q)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) { vector<int> ans; ans.reserve(queries.size()); for (const auto& q : queries) { const int rs = q[2] * q[2]; int cnt = 0; for (const auto& p : points) if ((q[0] - p[0]) * (q[0] - p[0]) + (q[1] - p[1]) * (q[1] - p[1]) <= rs) ++cnt; ans.push_back(cnt); } return ans; } }; |
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