LeetCode #1453 — HARD

Maximum Number of Darts Inside of a Circular Dartboard

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Alice is throwing n darts on a very large wall. You are given an array darts where darts[i] = [xi, yi] is the position of the ith dart that Alice threw on the wall.

Bob knows the positions of the n darts on the wall. He wants to place a dartboard of radius r on the wall so that the maximum number of darts that Alice throws lie on the dartboard.

Given the integer r, return the maximum number of darts that can lie on the dartboard.

Example 1:

Input: darts = [[-2,0],[2,0],[0,2],[0,-2]], r = 2
Output: 4
Explanation: Circle dartboard with center in (0,0) and radius = 2 contain all points.

Example 2:

Input: darts = [[-3,0],[3,0],[2,6],[5,4],[0,9],[7,8]], r = 5
Output: 5
Explanation: Circle dartboard with center in (0,4) and radius = 5 contain all points except the point (7,8).

Constraints:

  • 1 <= darts.length <= 100
  • darts[i].length == 2
  • -104 <= xi, yi <= 104
  • All the darts are unique
  • 1 <= r <= 5000

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Alice is throwing n darts on a very large wall. You are given an array darts where darts[i] = [xi, yi] is the position of the ith dart that Alice threw on the wall. Bob knows the positions of the n darts on the wall. He wants to place a dartboard of radius r on the wall so that the maximum number of darts that Alice throws lie on the dartboard. Given the integer r, return the maximum number of darts that can lie on the dartboard.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[-2,0],[2,0],[0,2],[0,-2]]
2

Example 2

[[-3,0],[3,0],[2,6],[5,4],[0,9],[7,8]]
5
Step 02

Core Insight

What unlocks the optimal approach

  • If there is an optimal solution, you can always move the circle so that two points lie on the boundary of the circle.
  • When the radius is fixed, you can find either 0 or 1 or 2 circles that pass two given points at the same time.
  • Loop for each pair of points and find the center of the circle, after that count the number of points inside the circle.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1453: Maximum Number of Darts Inside of a Circular Dartboard
class Solution {
    public int numPoints(int[][] darts, int r) {
        int n = darts.length;
        int maxDarts = 1;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                List<double[]> centers
                    = possibleCenters(darts[i][0], darts[i][1], darts[j][0], darts[j][1], r);
                for (double[] center : centers) {
                    maxDarts = Math.max(maxDarts, countDarts(center[0], center[1], darts, r));
                }
            }
        }
        return maxDarts;
    }

    private List<double[]> possibleCenters(int x1, int y1, int x2, int y2, int r) {
        List<double[]> centers = new ArrayList<>();
        double dx = x2 - x1;
        double dy = y2 - y1;
        double d = Math.sqrt(dx * dx + dy * dy);
        if (d > 2 * r) {
            return centers;
        }
        double midX = (x1 + x2) / 2.0;
        double midY = (y1 + y2) / 2.0;
        double distToCenter = Math.sqrt(r * r - (d / 2.0) * (d / 2.0));
        double offsetX = distToCenter * dy / d;
        double offsetY = distToCenter * -dx / d;

        centers.add(new double[] {midX + offsetX, midY + offsetY});
        centers.add(new double[] {midX - offsetX, midY - offsetY});
        return centers;
    }

    private int countDarts(double x, double y, int[][] darts, int r) {
        int count = 0;
        for (int[] dart : darts) {
            if (Math.sqrt(Math.pow(dart[0] - x, 2) + Math.pow(dart[1] - y, 2)) <= r + 1e-7) {
                count++;
            }
        }
        return count;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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