Find Four points such that they form a square whose sides are parallel to x and y axes in C++

C++Server Side ProgrammingProgramming

Concept

With respect of given ‘n’ pair of points, our task is to determine four points so that they form a square whose sides are parallel to x and y axes or else display “No such square”. It should be noted that if more than one square is possible then select the one with the maximum area.

Input

n = 6, points = (2, 2), (5, 5), (4, 5), (5, 4), (2, 5), (5, 2)

Output

Side of the square is: 3,
points of the square are 2, 2 5, 2 2, 5 5, 5

Explanation

The points 2, 2 5, 2 2, 5 5, 5 form a square of side 3

Input

n= 6, points= (2, 2), (5, 6), (4, 5), (5, 4), (8, 5), (4, 2)

Output

No such square

Method

Simple method − Select all possible pairs of points with four nested loops and then verifyif the points form a square which is parallel to principal axes. It has been seen that if yes then verify if it is the largest square so far in terms of the area and store the result, and then print the result at the end of the program.

Time Complexity − O(N^4)

Efficient method − Build a nested loop for top right and bottom left corner of the square and generate a square with those two points, then verify if the other two points which were assumed actually exist. Now to verify if a point exists or not, build a map and store the points in the map to decrease the time to verify whether the points exist.Moreover, keep in check the largest square by area so far and display it in the end.

Example

 Live Demo

// C++ implemenataion of the above approach
#include <bits/stdc++.h>
using namespace std;
// Determine the largest square
void findLargestSquare1(long long int points1[][2], int n1){
   // Used to map to store which points exist
   map<pair<long long int, long long int>, int> m1;
   // mark the available points
   for (int i = 0; i < n1; i++) {
      m1[make_pair(points1[i][0], points1[i][1])]++;
   }
   long long int side1 = -1, x1 = -1, y1 = -1;
   // Shows a nested loop to choose the opposite corners of square
   for (int i = 0; i < n1; i++) {
      // Used to remove the chosen point
      m1[make_pair(points1[i][0], points1[i][1])]--;
      for (int j = 0; j < n1; j++) {
         // Used to remove the chosen point
         m1[make_pair(points1[j][0], points1[j][1])]--;
         // Verify if the other two points exist
      if (i != j && (points1[i][0]-points1[j][0]) == (points1[i][1]-points1[j][1])){
         if (m1[make_pair(points1[i][0], points1[j][1])] > 0
            && m1[make_pair(points1[j][0], points1[i][1])] > 0) {
            // So if the square is largest then store it
         if (side1 < abs(points1[i][0] - points1[j][0])
            || (side1 == abs(points1[i][0] -points1[j][0])
            && ((points1[i][0] * points1[i][0]+ points1[i][1] * points1[i][1])
            < (x1 * x1 + y1 * y1)))) {
               x1 = points1[i][0];
               y1 = points1[i][1];
               side1 = abs(points1[i][0] - points1[j][0]);
            }
         }
      }
      // Used to add the removed point
      m1[make_pair(points1[j][0], points1[j][1])]++;
   }
   // Used to add the removed point
   m1[make_pair(points1[i][0], points1[i][1])]++;
}
// Used to display the largest square
if (side1 != -1)
   cout << "Side of the square is : " << side1
   << ", \npoints of the square are " << x1 << ", " << y1<< " "<< (x1 + side1) << ", " << y1
   << " "
   << (x1) << ", " << (y1 + side1)
   << " "
   << (x1 + side1) << ", " << (y1 + side1) << endl;
   else
   cout << "No such square" << endl;
}
//Driver code
int main(){
   int n1 = 6;
   // given points
   long long int points1[n1][2]= { { 2, 2 }, { 5, 5 }, { 4, 5 }, { 5, 4 }, { 2, 5 }, { 5, 2 }};
   // Determine the largest square
   findLargestSquare1(points1, n1);
   return 0;
}

Output

Side of the square is : 3,
points of the square are 2, 2 5, 2 2, 5 5, 5
raja
Published on 24-Jul-2020 14:40:08
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