Given N-sided polygon we need to find the number of triangles formed by joining the vertices of the given polygon with exactly one side being common.
Examples:
Input : 6
Output : 12
The image below is of a triangle forming inside a Hexagon by joining vertices as shown above. The two triangles formed has one side (AB) common with that of a polygon.It depicts that with one edge of a hexagon we can make two triangles with one side common. We can’t take C or F as our third vertex as it will make 2 sides common with the hexagon.
Triangle with one side common of a hexagon
No of triangles formed ie equal to 12 as there are 6 edges in a hexagon.Input : 5
Output : 5
Approach :
- To make a triangle with one side common with a polygon the two vertices adjacent to the chosen common vertices cannot be considered as the third vertex of a triangle.
- First select any one edge from the polygon. Consider this edge to be the common edge. Number of ways to select an edge in a polygon would be equal to n.
- Now ,to form a triangle ,select any of the (n-4) vertices left .Two vertices of the common edge and two vertices adjacent to the common edge cannot be considered.
- Number of triangle formed by joining the vertices of an n-sided polygon with one side common would be equal to n * ( n – 4) .
Below is the implementation of the above approach:
C++
// C++ program to implement // the above problem #include <bits/stdc++.h> using namespace std; // Function to find the number of triangles void findTriangles(int n) { int num; num = n * (n - 4); // print the number of triangles cout << num; } // Driver code int main() { // initialize the number of sides of a polygon int n; n = 6; findTriangles(n); return 0; } |
Java
// Java program to implement // the above problem class GFG { // Function to find the number of triangles static void findTriangles(int n) { int num; num = n * (n - 4); // print the number of triangles System.out.println(num); } // Driver code public static void main(String [] args) { // initialize the number of sides of a polygon int n; n = 6; findTriangles(n); } } // This code is contributed by ihritik |
Python3
# Python3 program to implement # the above problem # Function to find the number of triangles def findTriangles(n): num = n * (n - 4) # print the number of triangles print(num) # Driver code # initialize the number of sides of a polygon n = 6 findTriangles(n) # This codee is contributed by mohit kumar 29 |
C#
// C# program to implement // the above problem using System; class GFG { // Function to find the number of triangles static void findTriangles(int n) { int num; num = n * (n - 4); // print the number of triangles Console.WriteLine(num); } // Driver code public static void Main() { // initialize the number of sides of a polygon int n; n = 6; findTriangles(n); } } // This code is contributed by ihritik |
12
Time Complexity: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
