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 <bits/stdc++.h>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<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;}// Driver codeint main(){ Point arr[] = {{0, 0}, {1, 1}, {2, 0}, {2, 2}}; int n = sizeof(arr)/sizeof(arr[0]); cout << countPoints(arr, n); return 0;} |
Java
// Java program to count number // of triangles that can be formed // with given points in 2Dclass 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(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][0], Point[i][1], Point[j][0], Point[j][1], Point[k][0], Point[k][1])): 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
<?php// PHP program to count number // of triangles that can be formed // with given points in 2D// Returns determinant value // of three points in 2Dfunction det($x1, $y1, $x2, $y2, $x3, $y3){ return ($x1 * ($y2 - $y3) - $y1 * ($x2 - $x3) + 1 * ($x2 * $y3 - $y2 * $x3));}// Returns count of possible // triangles with given array// of points in 2D.function countPoints($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 = 0; $i < $n; $i++) for($j = $i + 1; $j < $n; $j++) for($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])) $result = $result + 1; return $result;}// Driver code$Point = array(array(0, 0), array(1, 1), array(2, 0), array(2, 2));$n = count($Point);echo countPoints($Point, $n);// This code is contributed by// mits?> |
Javascript
<script>// JavaScript program to count number // of triangles that can be formed // with given points in 2D// Returns determinant value // of three points in 2Dfunction 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.function countPoints(Point, n){ // Initialize result let result = 0; // Consider all triplets of // points given in inputs // Increment the result when // determinant of a triplet is not 0. for(let i = 0; i < n; i++) for(let j = i + 1; j < n; j++) for(let 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 Codelet Point = [ [ 0, 0 ], [ 1, 1 ], [ 2, 0 ], [ 2, 2 ] ];let n = Point.length;document.write(countPoints(Point, n));// This code is contributed by souravghosh0416</script> |
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.
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