Leetcode #1453: Maximum Number of Darts Inside of a Circular Dartboard
In this guide, we solve Leetcode #1453 Maximum Number of Darts Inside of a Circular Dartboard 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
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.
Quick Facts
- Difficulty: Hard
- 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: 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.
Python Solution
class Solution:
def numPoints(self, darts: list[list[int]], r: int) -> int:
def countDarts(x, y):
count = 0
for x1, y1 in darts:
if dist((x, y), (x1, y1)) <= r + 1e-7:
count += 1
return count
def possibleCenters(x1, y1, x2, y2):
dx, dy = x2 - x1, y2 - y1
d = sqrt(dx * dx + dy * dy)
if d > 2 * r:
return []
mid_x, mid_y = (x1 + x2) / 2, (y1 + y2) / 2
dist_to_center = sqrt(r * r - (d / 2) * (d / 2))
offset_x = dist_to_center * dy / d
offset_y = dist_to_center * -dx / d
return [
(mid_x + offset_x, mid_y + offset_y),
(mid_x - offset_x, mid_y - offset_y),
]
n = len(darts)
max_darts = 1
for i in range(n):
for j in range(i + 1, n):
centers = possibleCenters(
darts[i][0], darts[i][1], darts[j][0], darts[j][1]
)
for center in centers:
max_darts = max(max_darts, countDarts(center[0], center[1]))
return max_darts
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
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.