Find the unit place digit of sum of N factorials

Given a number N, the task is to find units place digit of the first N natural numbers factorials, i.e. 1!+2!+3!+….N! where N<=10e18.
Examples:

Input: n = 2 
Output: 3
1! + 2! = 3
Last digit is 3

Input: n = 3
Output: 9
1! + 2! + 3! = 9
Last digit is 9

Brute Force Approach:

In this approach, we are calculating the factorial of each number and then adding it to the sum. To prevent overflow, we are taking the modulo with 10 after every multiplication and addition operation. Finally, we return the unit place digit of the sum.

Below is implementation of the above approach:

C++

#include <iostream>
using namespace std;

int get_unit_digit(long long int N) {
    if (N == 0) {
        return 1;
    }
    long long int sum = 0;
    for (long long int i = 1; i <= N; i++) {
        long long int fact = 1;
        for (long long int j = 1; j <= i; j++) {
            fact *= j;
            fact %= 10; // to prevent overflow
        }
        sum += fact;
        sum %= 10; // to prevent overflow
    }
    return sum;
}

int main() {
    long long int N = 1;

    for (N = 0; N <= 10; N++)
        cout << "For N = " << N
             << " : " << get_unit_digit(N)
             << endl;

    return 0;
}

Output

For N = 0 : 1
For N = 1 : 1
For N = 2 : 3
For N = 3 : 9
For N = 4 : 3
For N = 5 : 3
For N = 6 : 3
For N = 7 : 3
For N = 8 : 3
For N = 9 : 3
For N = 10 : 3

Time Complexity: O(N ^ 2)

Auxiliary Space: O(1)

Efficient Approach: In this approach, only unit’s digit of N is to be calculated in the range [1, 5], because: 
1! = 1 
2! = 2 
3! = 6 
4! = 24 
5! = 120 
6! = 720 
7! = 5040 
so on.
As 5!=120, and factorial of number greater than 5 have trailing zeros. So, N>=5 doesn’t contribute in unit place while doing sum.
Therefore: 

if (n < 5)
    ans = (1 ! + 2 ! +..+ n !) % 10;
else
    ans = (1 ! + 2 ! + 3 ! + 4 !) % 10;

Note : We know (1! + 2! + 3! + 4!) % 10 = 3
So we always return 3 when n is greater 
than 4.

Algorithm :

Step 1: Start
Step 2: Start a static function called get_unit_digit which take integer value as its parameter called N.
Step 3: Now inside the above function set some conditions:
             a. If N is 0 or 1, return 1.
             b. If N is 2, return 3.
             c. If N is 3, return 9.
             d. If N is greater than or equal to 4, return 3.
Step 4: End

Below is the implementation of the efficient approach: 

C++

// C++ program to find the unit place digit
// of the first N natural numbers factorials
#include <iostream>
using namespace std;

// Function to find the unit's place digit
int get_unit_digit(long long int N)
{

    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
       return 1;
    else if (N == 2)
       return 3;
    else  if (N == 3)
       return 9;

    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
       return 3;
}

// Driver code
int main()
{
    long long int N = 1;

    for (N = 0; N <= 10; N++)
        cout << "For N = " << N
             << " : " << get_unit_digit(N)
             << endl;

    return 0;
}

Java

// Java  program to find the unit place digit
// of the first N natural numbers factorials

import java.io.*;

class GFG {


// Function to find the unit's place digit
static int get_unit_digit(  int N)
{

    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
    return 1;
    else if (N == 2)
    return 3;
    else if (N == 3)
    return 9;

    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
    return 3;
}

// Driver code

    public static void main (String[] args) {

      int N = 1;

    for (N = 0; N <= 10; N++)
            System.out.println ("For N = " + N
            + " : " + get_unit_digit(N));
    }
}
//This Code is Contributed by ajit

Python3

# Python3 program to find the unit
# place digit of the first N natural
# numbers factorials

# Function to find the unit's place digit
def get_unit_digit(N):

    # Let us write for cases when
    # N is smaller than or equal
    # to 4.
    if (N == 0 or N == 1):
        return 1
    elif (N == 2):
        return 3
    elif(N == 3):
        return 9

    # We know following
    # (1! + 2! + 3! + 4!) % 10 = 3
    else:
        return 3

# Driver code
N = 1
for N in range(11):
    print("For N = ", N, ":",
        get_unit_digit(N), sep = ' ')

# This code is contributed
# by sahilshelangia

C#

// C# program to find the unit
// place digit of the first N
// natural numbers factorials
using System;

class GFG
{

// Function to find the unit's
// place digit
static int get_unit_digit( int N)
{

    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
    return 1;
    else if (N == 2)
    return 3;
    else if (N == 3)
    return 9;

    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
    return 3;
}

// Driver code
static public void Main ()
{
    int N = 1;

    for (N = 0; N <= 10; N++)
        Console.WriteLine ("For N = " + N +
                " : " + get_unit_digit(N));
}
}

// This Code is Contributed by akt_mit

PHP

<?php
// PHP program to find the unit place digit
// of the first N natural numbers factorials

// Function to find the unit's place digit
function get_unit_digit($N)
{

    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if ($N == 0 || $N == 1)
        return 1;
    else if ($N == 2)
        return 3;
    else if ($N == 3)
        return 9;

    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
        return 3;
}

// Driver code
$N = 1;

for ($N = 0; $N <= 10; $N++)
    echo "For N = " . $N.
         " : " . get_unit_digit($N) . "\n";

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to find the unit place digit
// of the first N natural numbers factorials

// Function to find the unit's place digit
function get_unit_digit(N)
{

    // Let us write for cases when
    // N is smaller than or equal
    // to 4.
    if (N == 0 || N == 1)
      return 1;
    else if (N == 2)
      return 3;
    else if (N == 3)
      return 9;
    // We know following
    // (1! + 2! + 3! + 4!) % 10 = 3
    else // (N >= 4)
      return 3;
}

// Driver code
var N = 1;
for (N = 0; N <= 10; N++)
    document.write( "For N = " + N
        + " : " + get_unit_digit(N)+"<br>")


</script>

Output

For N = 0 : 1
For N = 1 : 1
For N = 2 : 3
For N = 3 : 9
For N = 4 : 3
For N = 5 : 3
For N = 6 : 3
For N = 7 : 3
For N = 8 : 3
For N = 9 : 3
For N = 10 : 3

Time Complexity: O(1)

Auxiliary Space: O(1)