Given an array S[] consisting of the lengths of the 6 sides of a Hexagon, the task is to calculate the number of equilateral triangles of unit length that can be made from the given hexagon.
Examples:
Approach:The following observations need to be made to solve the given problem:
 Consider an equilateral triangle of ‘X’ side length. It has been divided into smaller triangles of unit length each, by drawing lines parallel to its sides.
 Below are the images of three such equilateral triangles:
 In each of the above three examples, the count of unit length equilateral triangles possible are:
 X = 2: 4 equilateral triangles of 1 unit length side.
 X = 3: 9 equilateral triangles of 1 unit length side.
 X = 5: 25 equilateral triangles of 1 unit length side.
 By observation, it is clear that, for an equilateral triangle of side length X, X^{2} equilateral triangles of unit length are possible.
 Extending this observation to Hexagons, inscribe Hexagons inside the equilateral triangles, as shown below:
 It can be observed that by removing a certain number of mini triangles from the bigger triangle, the hexagon with given dimensions can be found.
The formula for counting the number of triangles of unit length can be generalized for a Hexagon having six sides S_{1 }, S_{2 }, S_{3 }, S_{4 }, S_{5 }, S_{6} as:
Number of triangles that can be formed = ( S_{1} + S_{2} + S_{3 })^{2} – S_{1}^{2} – S_{3}^{2} – S_{5}^{2}
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the // the number of Triangles possible int calculateTriangles( int sides[])
{ double count = pow (sides[0] + sides[1] +
sides[2], 2);
count = pow (sides[0], 2);
count = pow (sides[2], 2);
count = pow (sides[4], 2);
return ( int )(count);
} // Driver Code int main()
{ // Regular Hexagon
int sides[] = { 1, 1, 1, 1, 1, 1 };
cout << (calculateTriangles(sides)) << endl;
// Irregular Hexagon
int sides1[] = { 2, 2, 1, 3, 1, 2 };
cout << (calculateTriangles(sides1)) << endl;
return 0;
} // This code is contributed by 29AjayKumar 
// Java program to implement // the above approach import java.util.*;
class GFG{
// Function to calculate the // the number of Triangles possible static int calculateTriangles( int sides[])
{ double count = Math.pow(sides[ 0 ] + sides[ 1 ] +
sides[ 2 ], 2 );
count = Math.pow(sides[ 0 ], 2 );
count = Math.pow(sides[ 2 ], 2 );
count = Math.pow(sides[ 4 ], 2 );
return ( int )(count);
} // Driver Code public static void main(String[] args)
{ // Regular Hexagon
int sides[] = { 1 , 1 , 1 , 1 , 1 , 1 };
System.out.print((calculateTriangles(sides)) + "\n" );
// Irregular Hexagon
int sides1[] = { 2 , 2 , 1 , 3 , 1 , 2 };
System.out.print((calculateTriangles(sides1)) + "\n" );
} } // This code is contributed by amal kumar choubey 
# Python3 Program to implement # the above approach # Function to calculate the # the number of Triangles possible def calculateTriangles(sides):
count = pow ( sides[ 0 ] + sides[ 1 ] + sides[ 2 ], 2 )
count  = pow ( sides[ 0 ], 2 )
count  = pow ( sides[ 2 ], 2 )
count  = pow ( sides[ 4 ], 2 )
return int (count)
# Driver Code # Regular Hexagon sides = [ 1 , 1 , 1 , 1 , 1 , 1 ]
print (calculateTriangles(sides))
# Irregular Hexagon sides = [ 2 , 2 , 1 , 3 , 1 , 2 ]
print (calculateTriangles(sides))

// C# program to implement // the above approach using System;
class GFG{
// Function to calculate the // the number of Triangles possible static int calculateTriangles( int []sides)
{ double count = Math.Pow(sides[0] + sides[1] +
sides[2], 2);
count = Math.Pow(sides[0], 2);
count = Math.Pow(sides[2], 2);
count = Math.Pow(sides[4], 2);
return ( int )(count);
} // Driver Code public static void Main(String[] args)
{ // Regular Hexagon
int []sides = { 1, 1, 1, 1, 1, 1 };
Console.Write((calculateTriangles(sides)) + "\n" );
// Irregular Hexagon
int []sides1 = { 2, 2, 1, 3, 1, 2 };
Console.Write((calculateTriangles(sides1)) + "\n" );
} } // This code is contributed by amal kumar choubey 
6 19
Time Complexity: O(1)
Auxiliary Space: O(1)
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