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
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++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std; // This function returns the required number// of trianglesint countTriangles(pair<int, int> P[], int N){ // Hash Map to store the frequency of // slope corresponding to a point (X, Y) map<pair<int, int>, 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 functionint main(){ pair<int, int> P[] = { { 0, 0 }, { 2, 0 }, { 1, 1 }, { 2, 2 } }; int N = sizeof(P) / sizeof(P[0]); cout << countTriangles(P, N) << endl; return 0;} |
Python3
# Python3 implementation of the above approach from collections import defaultdictfrom 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 Codeif __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 |
3
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