Exponential factorial of N

 Given a positive integer N, the task is to print the Exponential factorial of N. Since the output can be very large, print the answer modulus 1000000007. 

Examples:  

Input: N = 4  
Output: 262144   

Input: N = 3  
Output: 9  

Approach: The given problem can be solved based on the following observations: 

The exponential factorial is defined by the recurrence relation:

• a_n=n^{a_{n-1}}                 .

Follow the steps below to solve the problem:

• Initialize a variable say res as 1 to store the exponential factorial of N.
• Iterate over the range [2, N] using the variable i and in each iteration update the res as res = i^res%1000000007.
• Finally, after completing the above step, print the answer obtained in res.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find exponential factorial
// of a given number
int ExpoFactorial(int N)
{
  
    // Stores the exponential factor of N
    int res = 1;
    int mod = 1000000007;

    // Iterate over the range [2, N]
    for (int i = 2; i < N + 1; i++)
      
        // Update res
        res = (int)pow(i, res) % mod;

    // Return res
    return res;
}

// Driver Code
int main()
{
    // Input
    int N = 4;
  
    // Function call
    cout << (ExpoFactorial(N));
  
   // This code is contributed by Potta Lokesh
    return 0;
}

// This code is contributed by lokesh potta

// Java program for the above approach
import java.io.*;
class GFG{
    
// Function to find exponential factorial
// of a given number
static int ExpoFactorial(int N)
{
    
    // Stores the exponential factor of N
    int res = 1;
    int mod = 1000000007;

    // Iterate over the range [2, N]
    for(int i = 2; i < N + 1; i++)
    
        // Update res
        res = (int)Math.pow(i, res) % mod;

    // Return res
    return res;
}

// Driver code
public static void main(String[] args)
{
    
    // Input
    int N = 4;

    // Function call
    System.out.println((ExpoFactorial(N)));
}
}

// This code is contributed by abhinavjain194

# Python3 program for the above approach

# Function to find exponential factorial
# of a given number


def ExpoFactorial(N):
    # Stores the exponential factor of N
    res = 1
    mod = (int)(1000000007)

    # Iterate over the range [2, N]
    for i in range(2, N + 1):
        # Update res
        res = pow(i, res, mod)

    # Return res
    return res


# Driver Code

# Input
N = 4
# Function call
print(ExpoFactorial(N))

// C# program for the above approach
using System;

class GFG{

// Function to find exponential factorial
// of a given number
static int ExpoFactorial(int N)
{
    
    // Stores the exponential factor of N
    int res = 1;
    int mod = 1000000007;

    // Iterate over the range [2, N]
    for(int i = 2; i < N + 1; i++)
    
        // Update res
        res = (int)Math.Pow(i, res) % mod;

    // Return res
    return res;
}

// Driver Code
public static void Main()
{
    // Input
    int N = 4;

    // Function call
    Console.Write(ExpoFactorial(N));

}
}

// This code is contributed by sanjoy_62.

<script>

// JavaScript program for the above approach


// Function to find exponential factorial
// of a given number
function ExpoFactorial(N) {

    // Stores the exponential factor of N
    let res = 1;
    let mod = 1000000007;

    // Iterate over the range [2, N]
    for (let i = 2; i < N + 1; i++)

        // Update res
        res = Math.pow(i, res) % mod;

    // Return res
    return res;
}

// Driver Code

// Input
let N = 4;

// Function call
document.write((ExpoFactorial(N)));

// This code is contributed by _saurabh_jaiswal

</script>

262144

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)