Leetcode #1828: Queries on Number of Points Inside a Circle
In this guide, we solve Leetcode #1828 Queries on Number of Points Inside a Circle in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Array, Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
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.
Python Solution
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
Complexity
The time complexity is , where and are the lengths of the arrays queries and points respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.