# Number of possible Triangles in a Cartesian coordinate system

• Difficulty Level : Easy
• Last Updated : 30 Aug, 2022

Given n points in a Cartesian coordinate system. Count the number of triangles formed.

Examples:

Input  : point[] = {(0, 0), (1, 1), (2, 0), (2, 2)
Output : 3
Three triangles can be formed from above points.

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.

## C++

 // C++ program to count number of triangles that can// be formed with given points in 2D#include using namespace std; // A point in 2Dstruct Point{   int x, y;}; // Returns determinant value of three points in 2Dint 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

## Java

 // Java program to count number// of triangles that can be formed// with given points in 2D class GFG{// Returns determinant value// of three points in 2Dstatic 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.static int countPoints(int [][]Point, 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(Point[i], Point[i],                    Point[j], Point[j],                    Point[k], Point[k])>=0)                    result = result + 1;     return result;} // Driver codepublic static void main(String[] args){int Point[][] = {{0, 0},{1, 1},{2, 0},{2, 2}};int n = Point.length;System.out.println(countPoints(Point, n));}}// This code is contributed by// mits

## Python3

 # Python program to count number# of triangles that can be formed# with given points in 2D # Returns determinant value# of three points in 2Ddef det(x1, y1, x2, y2, x3, 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.def countPoints(Point, n):     result = 0 # Initialize result     # Consider all triplets of    # points given in inputs    # Increment the result when    # determinant of a triplet is not 0.    for i in range(n):        for j in range(i + 1, n):            for k in range(j + 1, n):                if(det(Point[i], Point[i],                       Point[j], Point[j],                       Point[k], Point[k])):                    result = result + 1     return result # Driver codePoint = [[0, 0], [1, 1],         [2, 0], [2, 2]]n = len(Point)print(countPoints(Point, n)) # This code is contributed by# Sanjit_Prasad

## C#

 // C# program to count number// of triangles that can be formed// with given points in 2Dusing System; class GFG{// Returns determinant value// of three points in 2Dstatic 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.static int countPoints(int[,] Point, 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(Point[i,0], Point[i,1], Point[j,0], Point[j,1],Point[k,0], Point[k,1])>=0)                    result = result + 1;     return result;} // Driver codepublic static void Main(){int[,] Point = new int[,] { { 0, 0 }, { 1, 1 }, { 2, 0 }, { 2, 2 } };int n = Point.Length/Point.Rank;Console.WriteLine(countPoints(Point, n));}}// This code is contributed by mits

## PHP

 

## Javascript

 

Output:

3`

Time Complexity: .

Auxiliary Space: O(1) because it is using constant space

Optimization :
We can optimize the above solution to work in O(n2) using the fact that three 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.
This article is contributed by Vrushank Upadhyay. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up