# Find subfactorial of a number

• Last Updated : 08 Oct, 2021

Given an integer N, the task is to find the subfactorial of the number represented as !N. The subfactorial of a number is defined using below recurrence relation of a number N:

!N = (N-1) [ !(N-2) + !(N-1) ]

where !1 = 0 and !0 = 1

Some of the subfactorials are:

Examples:

Input: N = 4
Output: 9
Explanation:
!4 = !(4-1)*4 + (-1)4 = !3*4 + 1
!3 = !(3 – 1)*3 + (-1)3 = !2*3 – 1
!2 = !(2 – 1)*2 + (-1)2 = !1*2 + 1
!1 = !(1 – 1)*1 + (-1)1 = !0*1 – 1
Since !0 = 1, therefore !1 = 0, !2 = 1, !3 = 2 and !4 = 9.

Input: N = 0
Output: 1

Approach: The subfactorial of the number N can also be calculated as: Expanding this gives => !N = ( N! )*( 1 – 1/(1!) + (1/2!) – (1/3!)  …….. (1/N!)*(-1)N )

Therefore the above series can be used to find the subfactorial of number N. Follow the steps below to see how:

• Initialize variables, say res = 0, fact = 1 and count = 0.
• Iterate over the range from 1 to N using i and do the following:
• Update fact as fact*i.
• If the count is even then update res as res = res – (1 / fact).
• If the count is odd then update res as res = res + (1 / fact).
• Increase the value of count by 1.
• Finally, return fact*(1 + res).

Below is the implementation of the above approach:

## C++

 /// C++ program for the above approach#include using namespace std; // Function to find the subfactorial// of the numberdouble subfactorial(int N){     // Initialize variables    double res = 0, fact = 1;    int count = 0;     // Iterating over range N    for (int i = 1; i <= N; i++) {         // Fact variable store        // factorial of the i        fact = fact * i;         // If count is even        if (count % 2 == 0)            res = res - (1 / fact);        else            res = res + (1 / fact);         // Increase the value of        // count by 1        count++;    }     return fact * (1 + res);} // Driver Codeint main(){    int N = 4;    cout << subfactorial(N);     return 0;}

## Java

 /// Java program for the above approachimport java.util.*; class GFG {     // Function to find the subfactorial    // of the number    static double subfactorial(int N)    {         // Initialize variables        double res = 0, fact = 1;        int count = 0;         // Iterating over range N        for (int i = 1; i <= N; i++) {             // Fact variable store            // factorial of the i            fact = fact * i;             // If count is even            if (count % 2 == 0)                res = res - (1 / fact);            else                res = res + (1 / fact);             // Increase the value of            // count by 1            count++;        }         return fact * (1 + res);    }     // Driver Code    public static void main(String[] args)    {        int N = 4;        System.out.println((int)(subfactorial(N)));    }} // This code is contributed by ukasp.

## Python3

 # python program for the above approach # Function to find the subfactorial# of the numberdef subfactorial(N):         # Initialize variables    res = 0    fact = 1    count = 0     # Iterating over range N    for i in range(1, N+1):                 # Fact variable store                # factorial of the i        fact = fact * i         # If count is even        if (count % 2 == 0):            res = res - (1 / fact)         else:            res = res + (1 / fact)             # Increase the value of            # count by 1        count += 1     return fact * (1 + res) # Driver Codeif __name__ == "__main__":     N = 4    print(subfactorial(N))     # This code is contributed by rakeshsahni

## C#

 /// C# program for the above approachusing System;using System.Collections.Generic; class GFG{ // Function to find the subfactorial// of the numberstatic double subfactorial(int N){     // Initialize variables    double res = 0, fact = 1;    int count = 0;     // Iterating over range N    for (int i = 1; i <= N; i++) {         // Fact variable store        // factorial of the i        fact = fact * i;         // If count is even        if (count % 2 == 0)            res = res - (1 / fact);        else            res = res + (1 / fact);         // Increase the value of        // count by 1        count++;    }     return fact * (1 + res);} // Driver Codepublic static void Main(){    int N = 4;    Console.Write(subfactorial(N));}} // This code is contributed by ipg2016107.

## Javascript

 

Output
9

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

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