What is the brute force solution for Maximum Number of Darts Inside of a Circular Dartboard?

The brute force approach for Maximum Number of Darts Inside of a Circular Dartboard is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible circle that can be formed by pairs of darts. For each circle, we count how many darts fall within the radius. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Number of Darts Inside of a Circular Dartboard?

Maximum Number of Darts Inside of a Circular Dartboard uses the following concepts: Array, Math, Geometry. The recommended patterns to study are: Geometry, Brute Force, Two Pointers.

What are the interview tips for Maximum Number of Darts Inside of a Circular Dartboard?

Clearly explain your thought process and the geometric principles involved. Be mindful of edge cases, such as when all darts are far apart or when they are clustered tightly together. Practice visualizing the problem; drawing it out can help clarify your approach.

What are common mistakes when solving Maximum Number of Darts Inside of a Circular Dartboard?

Not considering the geometric properties of circles when calculating potential centers. Failing to account for cases where the distance between darts exceeds 2 * r, leading to unnecessary calculations.

#1453

Maximum Number of Darts Inside of a Circular Dartboard

Hard

ArrayMathGeometryGeometryBrute ForceTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

The optimal approach leverages the geometric properties of circles. By focusing on pairs of darts and calculating potential circle centers, we can efficiently determine how many darts can fit within the radius without redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: For each pair of darts, calculate the distance between them.
2Step 2: If the distance is less than or equal to 2 * r, calculate the two possible circle centers that can encompass both darts.
3Step 3: For each center, count how many darts are within the radius r.
4Step 4: Keep track of the maximum count of darts found.

solution.py22 lines

1# Full working Python code
2import math
3
4def maxDartsInside(darts, r):
5    def countDarts(center):
6        return sum(1 for x, y in darts if math.sqrt((x - center[0]) ** 2 + (y - center[1]) ** 2) <= r)
7
8    max_count = 0
9    n = len(darts)
10    for i in range(n):
11        for j in range(i + 1, n):
12            x1, y1 = darts[i]
13            x2, y2 = darts[j]
14            d = math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
15            if d > 2 * r:
16                continue
17            mid_x, mid_y = (x1 + x2) / 2, (y1 + y2) / 2
18            h = math.sqrt(r ** 2 - (d / 2) ** 2)
19            center1 = (mid_x + h * (y2 - y1) / d, mid_y - h * (x2 - x1) / d)
20            center2 = (mid_x - h * (y2 - y1) / d, mid_y + h * (x2 - x1) / d)
21            max_count = max(max_count, countDarts(center1), countDarts(center2))
22    return max_count

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops for pairs of darts, but we efficiently calculate potential circle centers and dart counts, minimizing unnecessary checks. The space complexity is O(1) as we only use a few variables.

1The maximum number of darts can be achieved by strategically placing the circle's center based on the positions of the darts.
2When the distance between two darts is less than or equal to 2 * r, it is possible to encompass both within a single circle.

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