Count the total number of triangles after Nth operation

Given an equilateral triangle, the task is to compute the total number of triangles after performing the following operation N times.
For every operation, the uncolored triangles are taken and divided into 4 equal equilateral triangles. Every inverted triangle formed is colored. Refer to the below figure for more details.

For N=1 the triangle formed is:

For N=2 the triangle formed is:

Examples:

Input :N = 10
Output : 118097

Input : N = 2
Output : 17

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

At every operation, 3 uncolored triangles, 1 colored triangle and the triangle itself is formed

On writing the above statement mathematically; count of triangles at Nth move = 3 * count of triangles at (N-1)th move + 2

Therefore, initializing a variable curr = 1 and tri_count = 0

Next, a loop is iterated from 1 to N

For every iteration, the operation mentioned above is performed.

Finally, the tri_count is returned

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h> 
using namespace std; 
// function to return the 
// total no.of Triangles 
int CountTriangles(int n) 
{ 
    int curr = 1; 
    int Tri_count = 0; 
    for (int i = 1; i <= n; i++) { 
        // For every subtriangle formed 
        // there are possibilities of 
        // generating (curr*3)+2 
  
        Tri_count = (curr * 3) + 2; 
        // Changing the curr value to Tri_count 
        curr = Tri_count; 
    } 
    return Tri_count; 
} 
  
// driver code 
int main() 
{ 
    int n = 10; 
    cout << CountTriangles(n); 
    return 0; 
} 

Java

class Gfg { 
    // Method to return the 
    // total no.of Triangles 
    public static int CountTriangles(int n) 
    { 
        int curr = 1; 
        int Tri_count = 0; 
        for (int i = 1; i <= n; i++) { 
            // For every subtriangle formed 
            // there are possibilities of 
            // generating (curr*3)+2 
  
            Tri_count = (curr * 3) + 2; 
            // Changing the curr value to Tri_count 
            curr = Tri_count; 
        } 
        return Tri_count; 
    } 
  
    // driver code 
    public static void main(String[] args) 
    { 
        int n = 10; 
        System.out.println(CountTriangles(n)); 
    } 
} 

Python

# Function to return the  
# total no.of Triangles 
def countTriangles(n): 
      
    curr = 1
    Tri_count = 0
    for i in range(1, n + 1): 
              
        # For every subtriangle formed 
        # there are possibilities of  
        # generating (curr * 3)+2 
        Tri_count = (curr * 3) + 2
        # Changing the curr value to Tri_count 
        curr = Tri_count 
    return Tri_count 
      
n = 10
print(countTriangles(n)) 

C#

using System; 
  
class Gfg  
{ 
    // Method to return the 
    // total no.of Triangles 
    public static int CountTriangles(int n) 
    { 
        int curr = 1; 
        int Tri_count = 0; 
        for (int i = 1; i <= n; i++)  
        { 
            // For every subtriangle formed 
            // there are possibilities of 
            // generating (curr*3)+2 
            Tri_count = (curr * 3) + 2; 
              
            // Changing the curr value to Tri_count 
            curr = Tri_count; 
        } 
        return Tri_count; 
    } 
  
    // Driver code 
    public static void Main(String[] args) 
    { 
        int n = 10; 
        Console.WriteLine(CountTriangles(n)); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

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