How do you solve Queries on Number of Points Inside a Circle on LeetCode?

Queries on Number of Points Inside a Circle (LeetCode #1828) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log n) time complexity and O(n) space complexity. Key insight: Understanding the geometry of circles and points is crucial.

What is the time complexity of Queries on Number of Points Inside a Circle?

The optimal solution for Queries on Number of Points Inside a Circle runs in O(n log n + m log n) time and uses O(n) space. The sorting step takes O(n log n), and each query takes O(log n) due to binary search, making it efficient for multiple queries.

What is the brute force solution for Queries on Number of Points Inside a Circle?

The brute force approach for Queries on Number of Points Inside a Circle is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each point against every query circle to see if the point lies within the circle. This is straightforward but can be slow for larger inputs. The optimal Optimal Solution approach improves this to O(n log n + m log n).

What algorithm or data structure is used to solve Queries on Number of Points Inside a Circle?

Queries on Number of Points Inside a Circle uses the following concepts: Array, Math, Geometry. The recommended patterns to study are: .

What are the interview tips for Queries on Number of Points Inside a Circle?

Clarify the problem statement and constraints with the interviewer. Discuss your thought process and approach before jumping into coding. Test your solution with edge cases and explain your reasoning.

What are common mistakes when solving Queries on Number of Points Inside a Circle?

Not considering points on the border of the circle. Confusing distance calculations (squared vs. actual distance).

#1828

Queries on Number of Points Inside a Circle

Medium

ArrayMathGeometry

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + m log n)Space O(n)

We can optimize the search by precomputing the distances of points from the origin and using a spatial data structure or sorting to quickly find how many points are within each query circle.

⚙️

Algorithm

4 steps

1Step 1: Create a list of squared distances for all points from the origin.
2Step 2: Sort this list of squared distances.
3Step 3: For each query, calculate the squared radius and use binary search to find how many points have squared distances less than or equal to this squared radius.
4Step 4: Store the results for each query and return the result list.

solution.py13 lines

1# Full working Python code
2from bisect import bisect_right
3
4def countPoints(points, queries):
5    distances = sorted((x**2 + y**2) for x, y in points)
6    result = []
7    for (xj, yj, rj) in queries:
8        r_squared = rj * rj
9        count = bisect_right(distances, r_squared)
10        result.append(count)
11    return result
12
13print(countPoints([[1,3],[3,3],[5,3],[2,2]], [[2,3,1],[4,3,1],[1,1,2]]))

ℹ

Complexity note: The sorting step takes O(n log n), and each query takes O(log n) due to binary search, making it efficient for multiple queries.

1Understanding the geometry of circles and points is crucial.
2Using sorting and binary search can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.