Maximum Number of Darts Inside of a Circular Dartboard

Maximum Number of Darts Inside of a Circular Dartboard - Problem

Circular Dartboard Optimization Challenge

Alice has thrown n darts at a massive wall, and you know the exact coordinates where each dart landed: darts[i] = [xi, yi] represents the position of the i-th dart.

Now Bob wants to strategically place a circular dartboard of radius r on the wall to maximize the number of Alice's darts that fall within or on the boundary of his dartboard.

Goal: Given the dart positions and the dartboard radius r, find the maximum number of darts that can be enclosed by optimally positioning the circular dartboard.

Note: A dart is considered inside the dartboard if the distance from the dart to the center of the dartboard is less than or equal to r.

Input & Output

example_1.py — Basic Square Pattern

$ Input: darts = [[-2,0],[2,0],[0,2],[0,-2]], r = 2

› Output: 4

💡 Note: A circle of radius 2 centered at origin (0,0) can cover all 4 darts positioned at the cardinal directions. Each dart is exactly distance 2 from the center, so they lie on the boundary of the circle.

example_2.py — Scattered Points

$ Input: darts = [[-3,0],[3,0],[2,6],[5,4],[0,9],[7,8]], r = 5

› Output: 5

💡 Note: The optimal dartboard placement can cover at most 5 darts. By carefully positioning the circle, we can include most points while staying within the radius constraint.

example_3.py — Single Point Edge Case

$ Input: darts = [[0,0]], r = 1

› Output: 1

💡 Note: With only one dart, the maximum possible coverage is 1. The dartboard can be placed anywhere that includes this single dart.

Visualization

Tap to expand

Understanding the Visualization

Brute Force Attempt

Try placing spotlight center at each target location - simple but misses optimal positions

Key Insight

Optimal spotlight position will always have at least 2 targets on its edge - this constrains possible positions

Geometric Solution

For each pair of targets, calculate all possible spotlight centers that put both targets on the edge

Count and Compare

For each valid spotlight position, count illuminated targets and track the maximum

Key Takeaway

🎯 Key Insight: The optimal dartboard position always has at least 2 darts on its circumference, dramatically reducing the search space from infinite positions to O(n²) candidates.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n darts, we check distance to all other n darts

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track maximum count and temporary calculations

✓ Linear Space

Constraints

1 ≤ darts.length ≤ 100
darts[i].length == 2
-10⁴ ≤ x_i, y_i ≤ 10⁴
1 ≤ r ≤ 5000
All dart coordinates are distinct
Answers within 10^-5 of the expected value are accepted

Asked in

G Google 23 a Amazon 18 f Meta 15 Apple 12

The optimal solution uses the geometric insight that any optimal circle must have at least 2 darts on its boundary. By checking all possible circles of radius r that pass through pairs of darts, we guarantee finding the maximum coverage. The algorithm runs in O(n³) time: O(n²) to consider all pairs, and O(n) to count darts in each circle.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try Every Dart as Center)	O(n²)	O(1)	Try placing dartboard center at each dart position and count enclosed darts
Circle Through Two Points	O(n³)	O(1)	Find all possible circles passing through pairs of darts, then count enclosed darts

Brute Force (Try Every Dart as Center) — Algorithm Steps

For each dart position, use it as a potential dartboard center
Count all darts within distance r from this center
Keep track of the maximum count found
Also try placing dartboard to include single dart (edge case)

Visualization

Tap to expand

Step-by-Step Walkthrough

Select Center

Choose a dart as potential dartboard center

Draw Circle

Draw circle of radius r around the chosen center

Count Darts

Count all darts within or on the circle boundary

Repeat

Try next dart as center and compare results

Code -

solution.c — C

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

int numPointsInCircle(int darts[][2], int n, int r) {
    if (n <= 1) return n;
    
    int maxCount = 1;
    
    // Try each dart as center
    for (int i = 0; i < n; i++) {
        double cx = darts[i][0], cy = darts[i][1];
        int count = 0;
        for (int j = 0; j < n; j++) {
            double dx = darts[j][0], dy = darts[j][1];
            double dist = sqrt((cx - dx) * (cx - dx) + (cy - dy) * (cy - dy));
            if (dist <= r + 1e-7) {
                count++;
            }
        }
        if (count > maxCount) {
            maxCount = count;
        }
    }
    
    return maxCount;
}

int main() {
    int darts[][2] = {{-2,0},{2,0},{0,2},{0,-2}};
    int r = 2;
    printf("%d\n", numPointsInCircle(darts, 4, r));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n darts, we check distance to all other n darts

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track maximum count and temporary calculations

✓ Linear Space

Constraints

1 ≤ darts.length ≤ 100
darts[i].length == 2
-10⁴ ≤ x_i, y_i ≤ 10⁴
1 ≤ r ≤ 5000
All dart coordinates are distinct
Answers within 10^-5 of the expected value are accepted

28.2K Views

Medium Frequency

~35 min Avg. Time

856 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try Every Dart as Center) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

Select Compiler