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;} |
Java
import java.util.*;class GFG { public static int get_unit_digit(long N) { if (N == 0) { return 1; } long sum = 0; for (long i = 1; i <= N; i++) { long fact = 1; for (long j = 1; j <= i; j++) { fact *= j; fact %= 10; // to prevent overflow } sum += fact; sum %= 10; // to prevent overflow } return (int)sum; } public static void main(String[] args) { long N; for (N = 0; N <= 10; N++) { System.out.println("For N = " + N + " : " + get_unit_digit(N)); } }} |
Python3
# Python code for the above approach# Function to get the unit digit of sum of# N factorialsdef get_unit_digit(N): # Base case: If N is 0 # The sum is 1 (as 0! = 1) if N == 0: return 1 sum = 0 # Traversing through all numbers from 1 to N for i in range(1, N + 1): fact = 1 # Calculating the factorial of 'i' for j in range(1, i + 1): fact *= j fact %= 10 # to prevent overflow sum += fact sum %= 10 # to prevent overflow return sumif __name__ == "__main__": for N in range(11): print(f"For N = {N} : {get_unit_digit(N)}") # This code is contributed by Pushpesh Raj |
C#
using System;class Program { // Function to get the unit digit of a number static int GetUnitDigit(long N) { if (N == 0) { return 1; } long sum = 0; for (long i = 1; i <= N; i++) { long fact = 1; for (long j = 1; j <= i; j++) { fact *= j; fact %= 10; // to prevent overflow } sum += fact; sum %= 10; // to prevent overflow } return (int)sum; } static void Main(string[] args) { long N = 1; for (N = 0; N <= 10; N++) Console.WriteLine("For N = " + N + " : " + GetUnitDigit(N)); }} |
Javascript
function get_unit_digit(N) { if (N === 0) { return 1; } let sum = 0; for (let i = 1; i <= N; i++) { let fact = 1; for (let j = 1; j <= i; j++) { fact *= j; fact %= 10; // to prevent overflow } sum += fact; sum %= 10; // to prevent overflow } return sum;}let N = 1;for (N = 0; N <= 10; N++) console.log("For N = " + N + " : " + get_unit_digit(N)); |
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 digitint 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 codeint 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 factorialsimport 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 digitdef 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 codeN = 1for 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 factorialsusing 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 |
Javascript
<script>// Javascript program to find the unit place digit// of the first N natural numbers factorials// Function to find the unit's place digitfunction 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 codevar N = 1;for (N = 0; N <= 10; N++) document.write( "For N = " + N + " : " + get_unit_digit(N)+"<br>") </script> |
PHP
<?php // PHP program to find the unit place digit// of the first N natural numbers factorials// Function to find the unit's place digitfunction 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?> |
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
