花花酱 LeetCode 1453. Maximum Number of Darts Inside of a Circular Dartboard

zxi — Sun, 17 May 2020 05:22:56 +0000

You have a very large square wall and a circular dartboard placed on the wall. You have been challenged to throw darts into the board blindfolded. Darts thrown at the wall are represented as an array of points on a 2D plane.

Return the maximum number of points that are within or lie on any circular dartboard of radius r.

Example 1:

Input: points = [[-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: points = [[-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).

Example 3:

Input: points = [[-2,0],[2,0],[0,2],[0,-2]], r = 1
Output: 1

Example 4:

Input: points = [[1,2],[3,5],[1,-1],[2,3],[4,1],[1,3]], r = 2
Output: 4

Constraints:

1 <= points.length <= 100
points[i].length == 2
-10^4 <= points[i][0], points[i][1] <= 10^4
1 <= r <= 5000

Solution 1: Angular Sweep

See for more details: https://www.geeksforgeeks.org/angular-sweep-maximum-points-can-enclosed-circle-given-radius/

Time complexity: O(n^2*logn)
Space complexity: O(n^2)

C++

// Author: Huahua
// Python-Like enumerate() In C++17
// http://reedbeta.com/blog/python-like-enumerate-in-cpp17/
template ())),
          typename = decltype(std::end(std::declval()))>
constexpr auto enumerate(T && iterable) {
  struct iterator {
    size_t i;
    TIter iter;
    bool operator != (const iterator & other) const { return iter != other.iter; }
    void operator ++ () { ++i; ++iter; }
    auto operator * () const { return std::tie(i, *iter); }
  };
  struct iterable_wrapper {
    T iterable;
    auto begin() { return iterator{ 0, std::begin(iterable) }; }
    auto end() { return iterator{ 0, std::end(iterable) }; }
  };
  return iterable_wrapper{ std::forward(iterable) };
}

class Solution {
public:
  int numPoints(vector>& points, int r) {
    const int n = points.size();
    vector> d(n, vector(n));
    
    for (const auto& [i, pi] : enumerate(points))  
      for (const auto& [j, pj] : enumerate(points))
        d[i][j] = d[j][i] = sqrt((pi[0] - pj[0]) * (pi[0] - pj[0]) 
                               + (pi[1] - pj[1]) * (pi[1] - pj[1]));
        
    int ans = 1;
    for (const auto& [i, pi] : enumerate(points)) {    
      vector> angles; // {angle, event_type}
      for (const auto& [j, pj] : enumerate(points)) {
        if (i != j && d[i][j] <= 2 * r) {
          const double a = atan2(pj[1] - pi[1], pj[0] - pi[0]);
          const double b = acos(d[i][j] / (2 * r));
          angles.emplace_back(a - b, -1); // entering
          angles.emplace_back(a + b, 1); // exiting
        }
      }
      // If angles are the same, entering points go first.
      sort(begin(angles), end(angles));
      int pts = 1;
      for (const auto& [_, event] : angles)
        ans = max(ans, pts -= event);
    }
    return ans;
  }
};

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花花酱 LeetCode 1453. Maximum Number of Darts Inside of a Circular Dartboard

Solution 1: Angular Sweep

C++