Find sum of factorials till N factorial (1! + 2! + 3! + … + N!)
Given a positive integer N. The task is to compute the sum of factorial from 1! to N!, 1! + 2! + 3! + … + N!.
Examples:
Input: N = 5
Output: 153
Explanation: 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153.
Input: N = 1
Output: 1
Naive Approach: The basic way to solve this problem is to find the factorial of all numbers till 1 to N and calculate their sum.
C++
#include<bits/stdc++.h>
using namespace std;
int findFact( int num) {
if (num == 0 || num == 1)
return 1;
long long result = 1;
for ( int i = 2; i <= num; i++) {
result *= i;
}
return result;
}
int findFactSum( int N) {
int sum = 0;
for ( int i = 1; i <= N; i++) {
sum += findFact(i);
}
return sum;
}
int main() {
int N = 5;
cout << findFactSum(N) << endl;
return 0;
}
|
Java
public class Main {
public static long findFact( int num) {
if (num == 0 || num == 1 )
return 1 ;
long result = 1 ;
for ( int i = 2 ; i <= num; i++) {
result *= i;
}
return result;
}
public static long findFactSum( int N) {
long sum = 0 ;
for ( int i = 1 ; i <= N; i++) {
sum += findFact(i);
}
return sum;
}
public static void main(String[] args) {
int N = 5 ;
System.out.println(findFactSum(N));
}
}
|
C#
using System;
class Program
{
static long FindFact( int num)
{
if (num == 0 || num == 1)
return 1;
long result = 1;
for ( int i = 2; i <= num; i++)
{
result *= i;
}
return result;
}
static long FindFactSum( int N)
{
long sum = 0;
for ( int i = 1; i <= N; i++)
{
sum += FindFact(i);
}
return sum;
}
static void Main( string [] args)
{
int N = 5;
Console.WriteLine(FindFactSum(N));
}
}
|
Javascript
function findFact(num) {
if (num === 0 || num === 1)
return 1;
let result = 1;
for (let i = 2; i <= num; i++) {
result *= i;
}
return result;
}
function findFactSum(N) {
let sum = 0;
for (let i = 1; i <= N; i++) {
sum += findFact(i);
}
return sum;
}
let N = 5;
console.log(findFactSum(N));
|
Python3
def find_fact(num):
if num = = 0 or num = = 1 :
return 1
result = 1
for i in range ( 2 , num + 1 ):
result * = i
return result
def find_fact_sum(N):
sum_result = 0
for i in range ( 1 , N + 1 ):
sum_result + = find_fact(i)
return sum_result
N = 5
print (find_fact_sum(N))
|
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Approach: An efficient approach is to calculate factorial and sum in the same loop making the time O(N). Traverse the numbers from 1 to N and for each number i:
- Multiply i with previous factorial (initially 1).
- Add this new factorial to a collective sum
At the end, print this collective sum. Below is the implementation of the above approach.
C++
#include <iostream>
using namespace std;
int findFactSum( int N)
{
int f = 1, Sum = 0;
for ( int i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
return Sum;
}
int main()
{
int N = 5;
cout << findFactSum(N);
return 0;
}
|
Java
class GFG {
static int findFactSum( int N)
{
int f = 1 , Sum = 0 ;
for ( int i = 1 ; i <= N; i++) {
f = f * i;
Sum += f;
}
return Sum;
}
public static void main(String[] args)
{
int N = 5 ;
System.out.print(findFactSum(N));
}
}
|
C#
using System;
class GFG
{
static int findFactSum( int N)
{
int f = 1, Sum = 0;
for ( int i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
return Sum;
}
public static void Main()
{
int N = 5;
Console.Write(findFactSum(N));
}
}
|
Javascript
<script>
function findFactSum(N)
{
let f = 1, Sum = 0;
for (let i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
return Sum;
}
let N = 5;
document.write(findFactSum(N));
</script>
|
Python3
def findFactSum(N):
f = 1
Sum = 0
for i in range ( 1 , N + 1 ):
f = f * i
Sum + = f
return Sum
if __name__ = = "__main__" :
N = 5
print (findFactSum(N))
|
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.
Last Updated :
29 Feb, 2024
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