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: 262144Input: 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:
.
.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 numberint 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 Codeint 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 approachimport java.io.*;class GFG{ // Function to find exponential factorial// of a given numberstatic 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 codepublic 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 numberdef 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# InputN = 4# Function callprint(ExpoFactorial(N)) |
C#
// C# program for the above approachusing System;class GFG{// Function to find exponential factorial// of a given numberstatic 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 Codepublic 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 numberfunction 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// Inputlet N = 4;// Function calldocument.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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