Count of Equilateral Triangles of unit length possible from a given Hexagon

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:

Input: S = {1, 1, 1, 1, 1, 1}
Output: 6
Explanation:

Input: S = {2, 2, 1, 3, 1, 2}
Output: 19
Explanation:

Approach:The following observations need to be made to solve the given problem:



The formula for counting the number of triangles of unit length can be generalized for a Hexagon having six sides S1 , S2 , S3 , S4 , S5 , S6 as:

Number of triangles that can be formed = ( S1 + S2 + S3 )2 – S12 – S32 – S52

Below is the implementation of the above approach:

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// 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
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// 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
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# 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))
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// 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 
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Output:
6
19

Time Complexity: O(1)
Auxiliary Space: O(1)

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