Number of triangles after N moves

Find the number of triangles in Nth step,
Rules: Draw an equilateral triangle at the start. In the i-th move, take the uncolored triangles, divides each of them in 4 parts of equal areas and color the central part. Keep a count of triangles till the Nth step.

Examples:

Input : 1
Output : 5 
Explanation: In 1st move we get


Input : 2
Output : 17 
Explanation: In 2nd move we get

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

Naive approach:
The number of triangles in the nth figure are 3 times the number of triangles in the (n-1)th figure+2. We can see by observation the nth figure is made by placing 3 triangles similar to that in (n-1) figure and an inverted triangle. We also take into account the bigger triangle that has been formed. Hence the number of triangles in the nth figure becomes (number of triangles in the (n-1)th figure)*3 + 2.

C++

// cpp program to calculate the number of equilateral 
// triangles 
#include <bits/stdc++.h> 
using namespace std; 
// fucntion to calculate number of traingles in Nth step 
int numberOfTriangles(int n) 
{ 
    int answer[n + 1] = { 0 }; 
    answer[0] = 1; 
    for (int i = 1; i <= n; i++)  
        answer[i] = answer[i - 1] * 3 + 2; 
      
    return answer[n]; 
} 
  
// driver program  
int main() 
{ 
    int n = 2; 
    cout << numberOfTriangles(n); 
    return 0; 
} 

Java

// Java program to find middle of three  
// distinct numbers to calculate the  
// number of equilateral triangles 
import java.util.*; 
  
class Triangle 
{    
    // fucntion to calculate number of  
    // traingles in Nth step 
    public static int numberOfTriangles(int n) 
    { 
        int[] answer = new int[n+1]; 
        answer[0] = 1; 
          
        for (int i = 1; i <= n; i++)  
            answer[i] = answer[i - 1] * 3 + 2; 
      
        return answer[n]; 
    } 
      
    // driver code 
    public static void main(String[] args) 
    { 
        int n = 2; 
        System.out.println(numberOfTriangles(n)); 
    } 
} 
  
// This code is contributed by rishabh_jain 

Python3

# Python3 code to calculate the  
# number of equilateral triangles 
  
# fucntion to calculate number  
# of traingles in Nth step 
def numberOfTriangles (n) : 
    answer = [None] * (n + 1); 
    answer[0] = 1; 
    i = 1
    while i <= n:  
        answer[i] = answer[i - 1] * 3 + 2; 
        i = i + 1
      
    return answer[n]; 
  
# Driver code 
n = 2
print(numberOfTriangles(n)) 
  
# This code is contributed by "rishabh_jain". 

C#

// C# program to find middle of three  
// distinct numbers to calculate the  
// number of equilateral triangles 
using System; 
  
class Triangle 
{  
    // fucntion to calculate number of  
    // traingles in Nth step 
    public static int numberOfTriangles(int n) 
    { 
        int[] answer = new int[n+1]; 
        answer[0] = 1; 
          
        for (int i = 1; i <= n; i++)  
            answer[i] = answer[i - 1] * 3 + 2; 
      
        return answer[n]; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int n = 2; 
        Console.WriteLine(numberOfTriangles(n)); 
    } 
} 
  
// This code is contributed by vt_m 

PHP

<?php 
// PHP program to calculate 
// the number of equilateral 
// triangles 
  
// function to calculate number 
// of traingles in Nth step 
function numberOfTriangles($n) 
{ 
    $answer = array(); 
    $answer[0] = 1; 
    for ($i = 1; $i <= $n; $i++)  
        $answer[$i] = $answer[$i - 1] *  
                               3 + 2; 
      
    return $answer[$n]; 
} 
  
    // Driver Code 
    $n = 2; 
    echo numberOfTriangles($n); 
  
// This code is contributed by anuj_67. 
?> 

Output:

Time Complexity: O(n)

An efficient solution will be to derive a formula for Nth step:
If we follow the naive approach for every step, then we get for Nth step the number of triangles to be

(2*(3^n))-1.

C++

// CPP program to calculate the number of  
// equilateral triangles 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to calculate number of triangles  
// in Nth step 
int numberOfTriangles(int n) 
{ 
    int ans = 2 * (pow(3, n)) - 1; 
    return ans; 
} 
  
// driver program  
int main() 
{ 
    int n = 2; 
    cout << numberOfTriangles(n); 
    return 0; 
} 

Java

// Java program to find middle of three  
// distinct numbers to calculate the 
// number of equilateral triangles 
import java.util.*; 
import static java.lang.Math.pow; 
  
class Triangle 
{    
    // function to calculate number  
    // of triangles in Nth step 
    public static double numberOfTriangles(int n) 
    { 
        double ans = 2 * (pow(3, n)) - 1; 
        return ans; 
    } 
      
    // driver code 
    public static void main(String[] args) 
    { 
        int n = 2; 
        System.out.println(numberOfTriangles(n)); 
    } 
} 
  
// This code is contributed by rishabh_jain 

Python3

# Python3 code to calculate the  
# number of equilateral triangles 
  
# fucntion to calculate number 
# of traingles in Nth step 
def numberOfTriangles (n) : 
    ans = 2 * (pow(3, n)) - 1; 
    return ans; 
  
# Driver code 
n = 2
print (numberOfTriangles(n)) 
  
# This code is contributed by "rishabh_jain". 

C#

//C# program to find middle of three  
// distinct numbers to calculate the 
// number of equilateral triangles 
using System; 
  
class Triangle 
{  
    // function to calculate number  
    // of triangles in Nth step 
    public static double numberOfTriangles(int n) 
    { 
        double ans = 2 * (Math.Pow(3, n)) - 1; 
        return ans; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int n = 2; 
        Console.WriteLine(numberOfTriangles(n)); 
    } 
} 
  
// This code is contributed by vt_m 

PHP

<?php 
// PHP program to calculate the  
// number of equilateral triangles 
  
// function to calculate  
// number of triangles  
// in Nth step 
function numberOfTriangles($n) 
{ 
    $ans = 2 * (pow(3, $n)) - 1; 
    return $ans; 
} 
  
    // Driver Code 
    $n = 2; 
    echo numberOfTriangles($n); 
  
// This code is contributed by anuj_67. 
?> 