Given a number n, we need to find the last digit in factorial n.
Input : n = 4
Output : 4
4! = 4 * 3 * 2 * 1. = 24. Last digit of 24 is 4.Input : n = 5
Output : 5
5! = 5*4 * 3 * 2 * 1. = 120. Last digit of 120 is 0.
A Naive Solution is to first compute fact = n!, then return the last digit of the result by doing fact % 10. This solution is inefficient and causes integer overflow for even slightly large value of n.
C++
// C++ program to find last digit in// factorial n using naive approach#include <iostream>using namespace std;int lastDigitFactorial(unsigned int n){ // Base case if (n == 0) return 1; // Calculate the factorial of n int fact = 1; for (int i = 1; i <= n; i++) { fact *= i; } // Return the last digit of the factorial return fact % 10;}// Driver codeint main() { cout<<lastDigitFactorial(6); return 0;} |
Java
public class Main { public static int lastDigitFactorial(int n) { // Base case if (n == 0) return 1; // Calculate the factorial of n int fact = 1; for (int i = 1; i <= n; i++) { fact *= i; } // Return the last digit of the factorial return fact % 10; } // Driver code public static void main(String[] args) { System.out.println(lastDigitFactorial(6)); }} |
Python3
# Python program to find last digit in# factorial n using naive approachdef last_digit_factorial(n): # Base case if n == 0: return 1 # Calculate the factorial of n fact = 1 for i in range(1, n + 1): fact *= i # Return the last digit of the factorial return fact % 10# Driver codeprint(last_digit_factorial(6)) |
C#
using System;public class GFG { public static int LastDigitFactorial(int n) { // Base case if (n == 0) return 1; // Calculate the factorial of n int fact = 1; for (int i = 1; i <= n; i++) { fact *= i; // To ensure that the factorial value doesn't overflow fact = fact % 10; } // Return the last digit of the factorial return fact % 10; } // Driver Code public static void Main(string[] args) { Console.WriteLine(LastDigitFactorial(6)); // This Code Is Contributed By Shubham Tiwari }} |
Javascript
// javascript code addition function lastDigitFactorial(n) { // Base case if (n === 0) return 1; // Calculate the factorial of n let fact = 1; for (let i = 1; i <= n; i++) { fact *= i; } // Return the last digit of the factorial return fact % 10;}// Driver codeconsole.log(lastDigitFactorial(6));// The code is contributed by Arushi Goel. |
Output
0
Time complexity:
Auxiliary space:
An Efficient Solution is based on the observation that all factorials after 5 have 0 as last digit.
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
…………..
C++
// C++ program to find last digit in// factorial n.#include <iostream>using namespace std;int lastDigitFactorial(unsigned int n){ // Explicitly handle all numbers // less than or equal to 4 if (n == 0) return 1; else if (n <= 2) return n; else if (n == 3) return 6; else if (n == 4) return 4; // For all numbers greater than 4 // the last digit is 0 else return 0;}int main() { cout<<lastDigitFactorial(6); return 0;} |
Java
// Java program to find last // digit in factorial n.import java.io.*; import java.util.*; class GFG { static int lastDigitFactorial(int n){ // Explicitly handle all numbers // less than or equal to 4 if (n == 0) return 1; else if (n <= 2) return n; else if (n == 3) return 6; else if (n == 4) return 4; // For all numbers greater than // 4 the last digit is 0 else return 0;}// Driver code public static void main(String[] args) { System.out.println(lastDigitFactorial(6));} } // This code is contributed by coder001 |
Python3
# Python3 program to find last digit in # factorial n.def lastDigitFactorial(n): # Explicitly handle all numbers # less than or equal to 4 if (n == 0): return 1 elif (n <= 2): return n elif (n == 3): return 6 elif (n == 4): return 4 # For all numbers greater than 4 # the last digit is 0 else: return 0print(lastDigitFactorial(6))# This code is contributed by divyeshrabadiya07 |
C#
// C# program to find last // digit in factorial n. using System;class GFG{ static int lastDigitFactorial(int n) { // Explicitly handle all numbers // less than or equal to 4 if (n == 0) return 1; else if (n <= 2) return n; else if (n == 3) return 6; else if (n == 4) return 4; // For all numbers greater than // 4 the last digit is 0 else return 0; } // Driver code public static void Main(string[] args) { Console.Write(lastDigitFactorial(6)); } } // This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript program to find last digit in // factorial n. function lastDigitFactorial(n) { // Explicitly handle all numbers // less than or equal to 4 if (n == 0) return 1; else if (n <= 2) return n; else if (n == 3) return 6; else if (n == 4) return 4; // For all numbers greater than 4 // the last digit is 0 else return 0; } // Driver code document.write(lastDigitFactorial(6)); // This code is contributed by Surbhi Tyagi</script> |
Output
0
Time Complexity: O(1), since no loop is used hence the algorithm takes constant time to execute
Auxiliary Space: O(1), since no extra array is used hence constant space is used
Approach 2: Using Loop:
The current approach works well for small values of n but is not efficient for large values of n. One way to optimize this approach is to compute the last digit of the factorial using modular arithmetic. The last digit of the factorial is the same as the last digit of the product of all the numbers from 1 to n.
Here’s the optimized approach:
C++
#include <iostream>using namespace std;int lastDigitFactorial(unsigned int n) { if (n == 0 || n == 1) { return 1; } int lastDigit = 1; for (int i = 2; i <= n; i++) { lastDigit = (lastDigit * i) % 10; } return lastDigit;}int main() { cout << lastDigitFactorial(6); return 0;} |
Java
public class Main { static int lastDigitFactorial(int n) { if (n == 0 || n == 1) { return 1; } int lastDigit = 1; for (int i = 2; i <= n; i++) { lastDigit = (lastDigit * i) % 10; } return lastDigit; } public static void main(String[] args) { System.out.println(lastDigitFactorial(6)); }} |
Python3
def lastDigitFactorial(n): if n == 0 or n == 1: return 1 lastDigit = 1 for i in range(2, n+1): lastDigit = (lastDigit * i) % 10 return lastDigitprint(lastDigitFactorial(6)) |
C#
using System;class Program { static int LastDigitFactorial(int n) { if (n == 0 || n == 1) { return 1; } int lastDigit = 1; for (int i = 2; i <= n; i++) { lastDigit = (lastDigit * i) % 10; } return lastDigit; } static void Main(string[] args) { Console.WriteLine(LastDigitFactorial(6)); }} |
Javascript
function lastDigitFactorial(n) { if (n == 0 || n == 1) { // if n is 0 or 1, return 1 return 1; } let lastDigit = 1; // initialize lastDigit to 1 for (let i = 2; i <= n; i++) { // loop from 2 to n // multiply lastDigit by i and take the last digit lastDigit = (lastDigit * i) % 10; } return lastDigit; // return the last digit of the factorial}// call the function with n = 6 and log the result to the consoleconsole.log(lastDigitFactorial(6)); |
Output:
0
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra array is used hence constant space is used
Now try below problems
1) Last non-zero digit in factorial
2) Count zeroes in factorial
3) First digit in Factorial
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