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Exponential factorial of N

  • Last Updated : 16 Jul, 2021

 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:  

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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 = ires%1000000007.
  • Finally, after completing the above step, print the answer obtained in res.

Below is the implementation of the above approach:

C++




// 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 exponetial factor of N
    int res = 1;
    int mod = 1000000007;
 
    // Iterare 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




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




# Python3 program for the above approach
 
# Function to find exponential factorial
# of a given number
 
 
def ExpoFactorial(N):
    # Stores the exponetial factor of N
    res = 1
    mod = (int)(1000000007)
 
    # Iterare 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#




// 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 exponetial factor of N
    int res = 1;
    int mod = 1000000007;
 
    // Iterare 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.

Javascript




<script>
 
// JavaScript program for the above approach
 
 
// Function to find exponential factorial
// of a given number
function ExpoFactorial(N) {
 
    // Stores the exponetial factor of N
    let res = 1;
    let mod = 1000000007;
 
    // Iterare 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>
Output
262144

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




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