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.
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));
}
}
|
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))
|
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>
|
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.