LeetCode 1828. Queries on Number of Points Inside a Circle
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)
?
Solution
class Solution {
public:
vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
vector<int> ans;
for (const vector<int>& query : queries) {
int num = 0;
int threshold = query[2] * query[2];
for (const vector<int>& point : points) {
int dis_square = (point[0] - query[0]) * (point[0] - query[0]) + (point[1] - query[1]) * (point[1] - query[1]);
if (dis_square <= threshold) {
num++;
}
}
ans.push_back(num);
}
return ans;
}
};
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