Random Point in Non-overlapping Rectangles in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of non-overlapping axis-aligned rectangles rects, we have to write a function pick which randomly and uniformly picks an integer number, point in the space covered by the rectangles. So we have to keep in mind some points −

  • An integer point is a point that has integer coordinates.
  • A point on the perimeter of a rectangle is included in the space covered by the rectangles.
  • The ith rectangle = rects[i] denotes [x1,y1,x2,y2], where [x1, y1] are the integer coordinates of the bottom-left corner, and [x2, y2] are the integer coordinates of the top-right corner.
  • The length and the width of each rectangle does not exceed 2000.
  • 1 <= rects.length <= 100
  • pick return a point as an array of integer coordinates [p_x, p_y]

If the input is like [1,1,5,5], and we call pick() three times, then the output will be [4,1], [4,1], [3,3]

To solve this, we will follow these steps −

  • Make two arrays area and rect
  • In the initializer do the following −
  • rect := rects, sum := 0
  • for i in range 0 to size of rects – 1
    • (x1, y1) := (rects[i, 0], rects[i, 1])
    • (x2, y2) := (rects[i, 2], rects[i, 3])
    • temp := |x2 – x1 + 1| * |y2 – y1 + 1|
    • sum := sum + temp, and insert sum into area
  • In the pick method, do the following −
  • randArea := random number mod sum + 1
  • for i in range 0 to size of area – 1
    • if randArea <= area[i], then come out from the loop
  • dist_x := random number mod |rect[i,0] – rect[i,2] + 1|
  • dist_y := random number mod |rect[i,1] – rect[i,3] + 1|
  • return a pair (dist_x + rect[i, 0], dist_y + rect[i, 1])

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
   public:
   vector <int> area;
   vector < vector <int> > rect;
   int sum;
   Solution(vector<vector<int> >& rects) {
      rect = rects;
      sum = 0;
      for(int i =0 ; i < rects.size(); i++){
         int x1 = rects[i][0];
         int y1 = rects[i][1];
         int x2 = rects[i][2];
         int y2 = rects[i][3];
         int temp = (abs(x2 - x1) + 1) * (abs(y2 - y1) + 1);
         sum += temp;
         area.push_back(sum);
      }
   }
   vector<int> pick() {
      int randArea = rand() % sum + 1;
      int i;
      for(i = 0; i < area.size(); i++){
         if(randArea <= area[i]) break;
      }
      int dist_x = rand() % (abs(rect[i][0] - rect[i][2] ) + 1);
      int dist_y = rand() % (abs(rect[i][1] - rect[i][3] ) + 1);
      return {dist_x + rect[i][0], dist_y + rect[i][1]};
   }
};
main(){
   vector<vector<int> > v = {{1, 1, 5, 5}};
   Solution ob(v);
   print_vector(ob.pick());
   print_vector(ob.pick());
   print_vector(ob.pick());
}

Input

["Solution", "pick", "pick", "pick"]
[[[[1, 1, 5, 5]]], [], [], []]

Output

[2, 3, ]
[4, 1, ]
[3, 5, ]
raja
Published on 02-May-2020 13:08:13
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