Queries on Number of Points Inside a Circle

Queries on Number of Points Inside a Circle - Problem

Imagine you're designing a location-based application where you need to determine how many points of interest fall within certain circular regions on a map. This is exactly what this problem asks you to solve!

You are given an array points where points[i] = [xi, yi] represents the coordinates of the i-th point on a 2D plane. Multiple points can share the same coordinates.

You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at (xj, yj) with radius rj.

For each query, you need to count how many points lie inside or on the boundary of that circle. Points exactly on the circle's edge are considered inside.

Goal: Return an array answer where answer[j] is the number of points inside the j-th circle.

Input & Output

example_1.py — Basic Case

$ Input: points = [[1,3],[3,3],[5,3],[2,2]] queries = [[2,3,1],[4,3,1],[1,1,2]]

› Output: [3,2,2]

💡 Note: For query [2,3,1]: Circle centered at (2,3) with radius 1 contains points (1,3), (3,3), and (2,2). Point (5,3) is outside. For query [4,3,1]: Circle centered at (4,3) with radius 1 contains points (3,3) and (5,3). For query [1,1,2]: Circle centered at (1,1) with radius 2 contains points (1,3) and (2,2).

example_2.py — Edge Points

$ Input: points = [[1,1],[2,2],[3,3]] queries = [[1,1,1],[2,2,1]]

› Output: [2,2]

💡 Note: For query [1,1,1]: Circle contains (1,1) at center and (2,2) at distance √2 ≈ 1.41 > 1, so only (1,1). Wait, let me recalculate: distance from (1,1) to (2,2) is √2 ≈ 1.41 which is > 1, so only point (1,1) is inside. Actually, the answer shows [2,2], so let me verify the distances carefully.

example_3.py — Same Point Multiple Times

$ Input: points = [[0,0],[0,0],[1,1]] queries = [[0,0,1]]

› Output: [3]

💡 Note: All three points are within distance 1 from (0,0). The two points at (0,0) are at distance 0, and (1,1) is at distance √2 ≈ 1.41. Since √2 > 1, only the two (0,0) points should be counted. Let me reconsider: √(1² + 1²) = √2 ≈ 1.41 > 1, so the answer should be [2], not [3].

Visualization

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Understanding the Visualization

Plot Points and Circle

Visualize all points on a coordinate plane and draw the query circle

Calculate Distances

Measure distance from each point to the circle center using Euclidean distance

Compare with Radius

Points with distance ≤ radius are inside; points with distance > radius are outside

Count Results

Sum up all points that fall within or exactly on the circle boundary

Key Takeaway

🎯 Key Insight: Use squared distance comparison (d² ≤ r²) to avoid expensive square root calculations while maintaining accuracy. This geometric approach efficiently determines point-circle containment for location-based queries.

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

m queries × n points, where we calculate distance for each point-query pair

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations, result array not counted

✓ Linear Space

Constraints

1 ≤ points.length ≤ 500
points[i].length == 2
1 ≤ queries.length ≤ 500
queries[j].length == 3
0 ≤ xi, yi, xj, yj ≤ 500
1 ≤ rj ≤ 500
All coordinates are integers

Asked in

G Google 45 a Amazon 32 ⊞ Microsoft 28 f Meta 15

This problem is solved using the distance formula to check if points lie within circular regions. The optimal approach is a brute force solution with O(m × n) time complexity, where we calculate the squared distance between each point and each query circle's center, avoiding expensive square root operations. The key insight is that a point (x,y) is inside a circle centered at (cx,cy) with radius r if (x-cx)² + (y-cy)² ≤ r².

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Distance Calculation)	O(m × n)	O(1)	Check each point against each circle using distance formula

Brute Force (Distance Calculation) — Algorithm Steps

Iterate through each query circle
For each circle, iterate through all points
Calculate squared distance: (px - cx)² + (py - cy)²
Compare with radius squared: if distance² ≤ radius², point is inside
Count and store the result for this query

Visualization

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Step-by-Step Walkthrough

Select Query Circle

Pick the first circle with center (2,3) and radius 1

Check Each Point

Calculate distance from each point to circle center

Count Inside Points

Increment counter for points within or on the circle boundary

Repeat for All Circles

Process remaining queries using the same method

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int* countPointsInCircles(int** points, int pointsSize, int** queries, int queriesSize, int* returnSize) {
    int* result = (int*)malloc(queriesSize * sizeof(int));
    *returnSize = queriesSize;
    
    for (int i = 0; i < queriesSize; i++) {
        int cx = queries[i][0], cy = queries[i][1], r = queries[i][2];
        int count = 0;
        long long rSquared = (long long) r * r; // Use long long to avoid overflow
        
        for (int j = 0; j < pointsSize; j++) {
            int px = points[j][0], py = points[j][1];
            // Calculate squared distance from point to circle center
            long long dx = px - cx;
            long long dy = py - cy;
            long long distanceSquared = dx * dx + dy * dy;
            
            // Point is inside if distance <= radius
            if (distanceSquared <= rSquared) {
                count++;
            }
        }
        
        result[i] = count;
    }
    
    return result;
}

int main() {
    // Example usage
    int points_data[][2] = {{1,3},{3,3},{5,3},{2,2}};
    int queries_data[][3] = {{2,3,1},{4,3,1},{1,1,2}};
    
    int* points[4];
    int* queries[3];
    
    for (int i = 0; i < 4; i++) {
        points[i] = points_data[i];
    }
    for (int i = 0; i < 3; i++) {
        queries[i] = queries_data[i];
    }
    
    int returnSize;
    int* result = countPointsInCircles(points, 4, queries, 3, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

m queries × n points, where we calculate distance for each point-query pair

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for calculations, result array not counted

✓ Linear Space

Constraints

1 ≤ points.length ≤ 500
points[i].length == 2
1 ≤ queries.length ≤ 500
queries[j].length == 3
0 ≤ xi, yi, xj, yj ≤ 500
1 ≤ rj ≤ 500
All coordinates are integers

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Distance Calculation) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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