Given a number, the task is to find the Superfactorial of a number. The result of multiplying the product of first n factorials is called Superfactorial of a number.
Superfactorial(n)= 1 ^ n * 2 ^ (n-1) * 3 ^ (n-2) * . . . . . * n ^ 1
Examples:
Input : 3
Output : 12 H(3) = 1! * 2! * 3! = 12Input : 4
Output : 288 H(4) = 1^4 * 2^3 * 3^2 * 4^1 = 288
An efficient approach is to compute all factorial iteratively till n, then compute the product of all factorial till n.
Python3
# Python3 program to find the# Superfactorial of a number# function to calculate the# value of Superfactorial def superfactorial(n): # initialise the # val to 1 val = 1 ans = [] for i in range(1, n + 1): val = val * i ans.append(val) # ans is the list with # all factorial till n. arr = [1] for i in range(1, len(ans)): arr.append((arr[-1]*ans[i])) return arr# Driver Coden = 5arr = superfactorial(n)print(arr[-1]) |
34560
Using built-in functions:
Time complexity: O(n)
Auxiliary Space: O(n)
Python3
import math# Function to calculate the Superfactorial of a numberdef superfactorial(n): # Initialize an empty list to store the factorials factorials = [] # Use a loop to calculate the factorial of each number till n for i in range(1, n+1): factorials.append(math.factorial(i)) # Initialize the result to 1 result = 1 # Use a loop to compute the product of all factorials for factorial in factorials: result *= factorial return result# Test the functionn = 5print(superfactorial(n))#This code is contributed by Edula Vinay Kumar Reddy |
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Explanation: Using built-in math.factorial() function, we can calculate the factorial of each number till n and then compute the product of all factorials using a loop. This method is useful when we have to do many mathematical operations on large dataset.
Time-complexity:O(N) Since super-factorials of number can be huge, hence the numbers will overflow. We can use boost libraries in C++ or BigInteger in Java to store the super-factorial of a number N.
Auxiliary Space: O(N)
Using Recursion
Python3
import mathdef superfactorial(n, acc=1): if n == 1: return acc else: return superfactorial(n-1, acc*math.factorial(n))# Driver coden = 5result = superfactorial(n)print(result)#this code is contributed by Jyothi Pinjala. |
34560
Time complexity: O(n)
Auxiliary Space: O(n)
