Number of possible Triangles in a Cartesian coordinate system - GeeksforGeeks

Number of possible Triangles in a Cartesian coordinate system - GeeksforGeeks
Given n points in a Cartesian coordinate system. Count the number of triangles formed.

A simple solution is to check if the determinant of the three points selected is non-zero or not. The following determinant gives the area of a Triangle (Also known as Cramer’s rule).

Area of the triangle with corners at (x1, y1), (x2, y2) and (x3, y3) is given by:

We can solve this by taking all possible combination of 3 points and finding the determinant.

// Returns determinant value of three points in 2D

int det(int x1, int y1, int x2, int y2, int x3, int y3)

{

   return x1*(y2 - y3) - y1*(x2 - x3) + 1*(x2*y3 - y2*x3);

}

 

// Returns count of possible triangles with given array

// of points in 2D.

int countPoints(Point arr[], int n)

{

    int result = 0;  // Initialize result

 

    // Consider all triplets of points given in inputs

    // Increment the result when determinant of a triplet

    // is not 0.

    for (int i=0; i<n; i++)

        for (int j=i+1; j<n; j++)

            for (int k=j+1; k<n; k++)

                if (det(arr[i].x, arr[i].y, arr[j].x,

                        arr[j].y, arr[k].x, arr[k].y))

                    result++;

 

    return result;

}

The complexity of the above solution code is O(n3).

Optimization : 
We can optimize above solution to work in O(n2) using the fact that hree points cannot form a triangle if they are collinear. We can use hashing to store slopes of all pairs and find all triangles in O(n2) time.

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