Perfect Rectangle in C++

C++Server Side ProgrammingProgramming

Suppose we have N axis-aligned rectangles, we have to check whether they all together form an exact cover of a rectangular region or not. Here each rectangle is represented as a bottom-left point and a top-right point. So a unit square is represented as [1,1,2,2]. (bottom-left point is (1, 1) and top-right point is (2, 2)).

So, if the input is like rectangles = [[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]], then the output will be true as all 5 rectangles together form an exact cover of a rectangular region.

To solve this, we will follow these steps −

  • Define one set visited

  • area := 0

  • x2 := -inf, x1 := inf

  • y2 := -inf, y1 := inf

  • for each r in given list re −

    • x1 := minimum of r[0] and x1

    • x2 := maximum of r[2] and x2

    • y1 := minimum of r[1] and y1

    • y2 := maximum of r[3] and y2

    • area := area + ((r[2] - r[0]) * (r[3] - r[1]))

    • s1 := r[0] concatenate r[1]

    • s2 := r[0] concatenate r[3]

    • s3 := r[2] concatenate r[3]

    • s4 := r[2] concatenate r[1]

    • if s1 is visited, then −

      • delete s1 from visited

    • Otherwise

      • insert s1 into visited

    • if s2 is visited, then −

      • delete s2 from visited

    • Otherwise

      • insert s2 into visited

    • if s3 is visited, then −

      • delete s3 from visited

    • Otherwise

      • insert s3 into visited

    • if s4 is visited, then −

      • delete s4 from visited

    • Otherwise

      • insert s4 into visited

  • s1 := concatenate x1 and y1

  • s2 := concatenate x2 and y1

  • s3 := concatenate x1 and y2

  • s4 := concatenate x2 and y2

  • if s1, s2, s3, s4 all are not visited, then

    • return false

  • return true when area is same as ((x2 - x1) * (y2 - y1))

Example  

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool isRectangleCover(vector<vector<int>> &re) {
      unordered_set<string> visited;
      int area = 0;
      int x2 = INT_MIN;
      int x1 = INT_MAX;
      int y2 = INT_MIN;
      int y1 = INT_MAX;
      for (auto &r : re) {
         x1 = min(r[0], x1);
         x2 = max(r[2], x2);
         y1 = min(r[1], y1);
         y2 = max(r[3], y2);
         area += (r[2] - r[0]) * (r[3] - r[1]);
         string s1 = to_string(r[0]) + to_string(r[1]);
         string s2 = to_string(r[0]) + to_string(r[3]);
         string s3 = to_string(r[2]) + to_string(r[3]);
         string s4 = to_string(r[2]) + to_string(r[1]);
         if (visited.count(s1)) {
            visited.erase(s1);
         }
         else
            visited.insert(s1);
         if (visited.count(s2)) {
            visited.erase(s2);
         }
         else
            visited.insert(s2);
         if (visited.count(s3)) {
            visited.erase(s3);
         }
         else
            visited.insert(s3);
         if (visited.count(s4)) {
            visited.erase(s4);
         }
         else
            visited.insert(s4);
         }
         string s1 = to_string(x1) + to_string(y1);
         string s2 = to_string(x2) + to_string(y1);
         string s3 = to_string(x1) + to_string(y2);
         string s4 = to_string(x2) + to_string(y2);
         if (!visited.count(s1) || !visited.count(s2) || !visited.count(s3) || !visited.count(s4) || visited.size() != 4)
            return false;
         return area == (x2 - x1) * (y2 - y1);
      }
};
main() {
   Solution ob;
   vector<vector<int>> v = {{1, 1, 3, 3}, {3, 1, 4, 2}, {3, 2, 4, 4}, {1, 3, 2, 4}, {2, 3, 3, 4}};
   cout << (ob.isRectangleCover(v));
}

Input

{{1, 1, 3, 3}, {3, 1, 4, 2}, {3, 2, 4, 4}, {1, 3, 2, 4}, {2, 3, 3, 4}}

Output

1
raja
Published on 21-Jul-2020 12:02:58
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