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.
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++ 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 program to implement // the above problem import java.io.*;
public 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 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 code is contributed by mohit kumar 29 |
// 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 |
<script> // JavaScript program to implement // the above problem // Function to find the number of triangles function findTriangles(n)
{ var num;
num = n * (n - 4);
// print the number of triangles
document.write(num);
} // Driver code // initialize the number of sides of a polygon var n;
n = 6; findTriangles(n); // This code contributed by shikhasingrajput </script> |
12
Time Complexity: O(1)
Auxiliary Space: O(1)