# Number of triangles that can be formed with given N points

Given X and Y coordinates of N points on a Cartesian plane. The task is to find the number of possible triangles with the non-zero area that can be formed by joining each point to every other point.

Examples:

Input: P[] = {{0, 0}, {2, 0}, {1, 1}, {2, 2}}
Output: 3
Possible triangles can be [(0, 0}, (2, 0), (1, 1)],
[(0, 0), (2, 0), (2, 2)] and [(1, 1), (2, 2), (2, 0)]

Input : P[] = {{0, 0}, {2, 0}, {1, 1}}
Output : 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Naive approach has been already discussed in Number of possible Triangles in a Cartesian coordinate system

Efficient Approach: Consider a point Z and find its slope with every other point. Now, if two points are having the same slope with point Z that means the 3 points are collinear and they cannot form a triangle. Hence, the number of triangles having Z as one of its points is the number of ways of choosing 2 points from the remaining points and then subtracting the number of ways of choosing 2 points from points having the same slope with Z. Since Z can be any point among N points, we have to iterate one more loop.

Below is the implementation of above approach:

 // C++ implementation of the above approach #include using namespace std;    // This function returns the required number // of triangles int countTriangles(pair P[], int N) {     // Hash Map to store the frequency of     // slope corresponding to a point (X, Y)     map, int> mp;     int ans = 0;        // Iterate over all possible points     for (int i = 0; i < N; i++) {         mp.clear();            // Calculate slope of all elements         // with current element         for (int j = i + 1; j < N; j++) {             int X = P[i].first - P[j].first;             int Y = P[i].second - P[j].second;                // find the slope with reduced             // fraction             int g = __gcd(X, Y);             X /= g;             Y /= g;             mp[{ X, Y }]++;         }         int num = N - (i + 1);            // Total number of ways to form a triangle         // having one point as current element         ans += (num * (num - 1)) / 2;            // Subtracting the total number of ways to         // form a triangle having the same slope or are         // collinear         for (auto j : mp)             ans -= (j.second * (j.second - 1)) / 2;     }     return ans; }    // Driver Code to test above function int main() {     pair P[] = { { 0, 0 }, { 2, 0 }, { 1, 1 }, { 2, 2 } };     int N = sizeof(P) / sizeof(P[0]);     cout << countTriangles(P, N) << endl;     return 0; }

 # Python3 implementation of the above approach  from collections import defaultdict from math import gcd    # This function returns the  # required number of triangles  def countTriangles(P, N):         # Hash Map to store the frequency of      # slope corresponding to a point (X, Y)      mp = defaultdict(lambda:0)      ans = 0        # Iterate over all possible points      for i in range(0, N):          mp.clear()             # Calculate slope of all elements          # with current element          for j in range(i + 1, N):              X = P[i][0] - P[j][0]              Y = P[i][1] - P[j][1]                 # find the slope with reduced              # fraction              g = gcd(X, Y)              X //= g              Y //= g              mp[(X, Y)] += 1                    num = N - (i + 1)             # Total number of ways to form a triangle          # having one point as current element          ans += (num * (num - 1)) // 2            # Subtracting the total number of          # ways to form a triangle having          # the same slope or are collinear          for j in mp:              ans -= (mp[j] * (mp[j] - 1)) // 2            return ans     # Driver Code if __name__ == "__main__":         P = [[0, 0], [2, 0], [1, 1], [2, 2]]      N = len(P)      print(countTriangles(P, N))        # This code is contributed by Rituraj Jain

Output:
3

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Improved By : rituraj_jain