LeetCode 1453: Maximum Number of Darts Inside of a Circular Dartboard Solution
Master LeetCode problem 1453 (Maximum Number of Darts Inside of a Circular Dartboard), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
To solve this problem, we can iterate through all possible pairs of darts and consider the circle with that pair as the diameter. For each circle, we can count how many other darts lie inside or on the boundary of the circle. The circle with the maximum count of darts inside or on the boundary will be the optimal circle for placing the dartboard.
To check if a dart lies inside a circle, we can calculate the distance between the center of the circle and the dart. If this distance is less than or equal to the radius of the circle, the dart lies inside or on the boundary of the circle.
Time Complexity:
The time complexity of this approach is O(n^3), where n is the number of darts.
Space Complexity:
The space complexity of this approach is O(1) as we are not using any extra space apart from a few variables.
:
Solution Code
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++) {
double[] center = {(darts[i][0] + darts[j][0]) / 2.0, (darts[i][1] + darts[j][1]) / 2.0};
double distance = Math.sqrt(Math.pow(darts[i][0] - center[0], 2) + Math.pow(darts[i][1] - center[1], 2));
if (distance > r) continue;
int count = 2;
for (int k = 0; k < n; k++) {
if (k != i && k != j) {
double dist = Math.sqrt(Math.pow(darts[k][0] - center[0], 2) + Math.pow(darts[k][1] - center[1], 2));
if (dist <= r) {
count++;
}
}
}
maxDarts = Math.max(maxDarts, count);
}
}
return maxDarts;
}
}Try It Yourself
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What is the time complexity of LeetCode 1453 (Maximum Number of Darts Inside of a Circular Dartboard)?
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