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:
def countPoints(self, points: List[List[int]], queries: List[List[int]]) -> List[int]:
answer = []
for x, y, r in queries:
numPoints = 0
for cx, cy in points:
if (cx - x) ** 2 + (cy - y) ** 2 <= r ** 2:
numPoints += 1
answer.append(numPoints)
return answer
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