We are given two numbers A and B such that B >= A. We need to compute the last digit of this resulting F such that F = B!/A! where 1 = A, B <= 10^18 (A and B are very large).
Examples:
Input : A = 2, B = 4
Output : 2
Explanation : A! = 2 and B! = 24.
F = 24/2 = 12 --> last digit = 2
Input : 107 109
Output : 2
As we know, factorial function grows on an exponential rate. Even the largest data type
cannot hold factorial of numbers like 100. To compute factorial of moderately large numbers, refer this.
Here the given constraints are very large. Thus, calculating the two factorials and later
dividing them and computing the last digit is practically an impossible task.
Thus we have to find an alternate approach to break down our problem. It is known that the last digit of factorial always belongs to the set {0, 1, 2, 4, 6}
The approach is as follows: –
1) We evaluate the difference between B and A
2) If the (B – A) >= 5, then the answer is always 0
3) If the difference (B – A) < 5, then we iterate from (A+1) to B, multiply and store them. multiplication_answer % 10 shall be our answer.
C++
// CPP program to find last digit of a number // obtained by dividing factorial of a number// with factorial of another number.#include <iostream>using namespace std;// Function which computes the last digit// of resultant of B!/A!int computeLastDigit(long long int A, long long int B){ int variable = 1; if (A == B) // If A = B, B! = A! and B!/A! = 1 return 1; // If difference (B - A) >= 5, answer = 0 else if ((B - A) >= 5) return 0; else { // If none of the conditions are true, we // iterate from A+1 to B and multiply them. // We are only concerned for the last digit, // thus we take modulus of 10 for (long long int i = A + 1; i <= B; i++) variable = (variable * (i % 10)); return variable % 10; }}// driver functionint main(){ cout << computeLastDigit(2632, 2634); return 0;} |
C
// CPP program to find last digit of a number // obtained by dividing factorial of a number// with factorial of another number.#include <stdio.h>// Function which computes the last digit// of resultant of B!/A!int computeLastDigit(long long int A, long long int B){ int variable = 1; if (A == B) return 1; // If difference (B - A) >= 5, answer = 0 else if ((B - A) >= 5) return 0; else { // If none of the conditions are true, we // iterate from A+1 to B and multiply them. // We are only concerned for the last digit, // thus we take modulus of 10 for (long long int i = A + 1; i <= B; i++) variable = (variable * (i % 10)); return variable % 10; }}// driver functionint main(){ long long int a=2632; long long int b=2634; int ans=computeLastDigit(a,b); printf("%d",ans); return 0;}// This code is contributed by allwink45. |
Java
// Java program to find last digit of a number // obtained by dividing factorial of a number// with factorial of another number.import java.io.*;class GFG {// Function which computes the last digit// of resultant of B!/A!static int computeLastDigit(long A, long B){ int variable = 1; if (A == B) // If A = B, B! = A! and B!/A! = 1 return 1; // If difference (B - A) >= 5, answer = 0 else if ((B - A) >= 5) return 0; else { // If none of the conditions are true, we // iterate from A+1 to B and multiply them. // We are only concerned for the last digit, // thus we take modulus of 10 for (long i = A + 1; i <= B; i++) variable = (int)(variable * (i % 10)) % 10; return variable % 10; }}// driver functionpublic static void main(String[] args){ System.out.println(computeLastDigit(2632, 2634));}}// This article is contributed by Prerna Saini |
Python3
# Python program to find# last digit of a number # obtained by dividing# factorial of a number# with factorial of another number.# Function which computes# the last digit# of resultant of B!/A!def computeLastDigit(A,B): variable = 1 if (A == B): # If A = B, B! = A! and B!/A! = 1 return 1 # If difference (B - A) >= 5, answer = 0 elif ((B - A) >= 5): return 0 else: # If none of the conditions # are true, we # iterate from A+1 to B # and multiply them. # We are only concerned # for the last digit, # thus we take modulus of 10 for i in range(A + 1, B + 1): variable = (variable * (i % 10)) % 10 return variable % 10 # driver functionprint(computeLastDigit(2632, 2634))# This code is contributed# by Anant Agarwal. |
C#
// C# program to find last digit of// a number obtained by dividing // factorial of a number with// factorial of another number.using System;class GFG {// Function which computes the last // digit of resultant of B!/A!static int computeLastDigit(long A, long B){ int variable = 1; // If A = B, B! = A! // and B!/A! = 1 if (A == B) return 1; // If difference (B - A) >= 5, // answer = 0 else if ((B - A) >= 5) return 0; else { // If none of the conditions are true, we // iterate from A+1 to B and multiply them. // We are only concerned for the last digit, // thus we take modulus of 10 for (long i = A + 1; i <= B; i++) variable = (int)(variable * (i % 10)) % 10; return variable % 10; }}// Driver Codepublic static void Main(){ Console.WriteLine(computeLastDigit(2632, 2634));}}// This code is contributed by vt_m. |
Javascript
<script>// Javascript program to find last digit of a number // obtained by dividing factorial of a number // with factorial of another number. // Function which computes the last digit // of resultant of B!/A! function computeLastDigit(A, B) { let variable = 1; if (A == B) // If A = B, B! = A! and B!/A! = 1 return 1; // If difference (B - A) >= 5, answer = 0 else if ((B - A) >= 5) return 0; else { // If none of the conditions are true, we // iterate from A+1 to B and multiply them. // We are only concerned for the last digit, // thus we take modulus of 10 for (let i = A + 1; i <= B; i++) variable = (variable * (i % 10)) % 10; return variable % 10; } } // driver function document.write(computeLastDigit(2632, 2634)); // This code is contributed by Surbhi Tyagi</script> |
PHP
<?php// PHP program to find last digit of a number // obtained by dividing factorial of a number// with factorial of another number.// Function which computes the last // digit of resultant of B!/A!function computeLastDigit($A, $B){ $variable = 1; // If A = B, B! = A! // and B!/A! = 1 if ($A == $B) return 1; // If difference (B - A) >= 5, // answer = 0 else if (($B - $A) >= 5) return 0; else { // If none of the conditions // are true, we iterate from // A+1 to B and multiply them. // We are only concerned for // the last digit, thus we // take modulus of 10 for ($i = $A + 1; $i <= $B; $i++) $variable = ($variable * ($i % 10)) % 10; return $variable % 10; }} // Driver Code echo computeLastDigit(2632, 2634);// This code is contributed by ajit?> |
Output:
2
Time Complexity: O(1).
Auxiliary Space: O( 1 ), since no extra space has been taken.
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Using Recursion :
Approach:
This approach calculates the factorial of A using recursion and divides the factorial of B with it to get the last digit.
Define a function factorial(n) that takes an integer n and calculates its factorial using recursion.
Define another function last_digit(A, B) that takes two integers A and B.
Calculate the factorials of A and B using the factorial() function defined earlier.
Divide the factorial of B by the factorial of A to get the quotient.
Return the last digit of the quotient by using the modulus operator % with the divisor as 10.
C++
#include <iostream>// Function to calculate the last digit of B! / A! without calculating factorialsint lastDigit(int A, int B) { int result = 1; // Calculate the last digit of each number from A+1 to B for (int i = A + 1; i <= B; ++i) { result *= (i % 10); // Keep only the last digit result %= 10; // Ensure result remains a single digit } return result;}int main() { // Example usage std::cout << lastDigit(2, 4) << std::endl; // Output: 2 std::cout << lastDigit(107, 109) << std::endl; // Output: 2 // This Code is Contributed by Shivam Tiwari return 0;} |
Java
// Java Code for the above approachpublic class GFG { // Function to calculate the last digit of B! / A! // without calculating factorials public static int lastDigit(int A, int B) { int result = 1; // Calculate the last digit of each number from A+1 // to B for (int i = A + 1; i <= B; ++i) { result *= (i % 10); // Keep only the last digit result %= 10; // Ensure result remains a single // digit } return result; } public static void main(String[] args) { // Example usage System.out.println(lastDigit(2, 4)); // Output: 2 System.out.println( lastDigit(107, 109)); // Output: 2 }}// This code is contributed by Susobhan Akhuli |
Python3
def factorial(n): if n == 0: return 1 return n * factorial(n-1)def last_digit(A, B): fact_A = factorial(A) fact_B = factorial(B) return fact_B // fact_A % 10# Example usageprint(last_digit(2, 4)) # Output: 2print(last_digit(107, 109)) # Output: 2 |
C#
using System;class Program{ // Function to calculate the last digit of B! / A! without calculating factorials static int LastDigit(int A, int B) { int result = 1; // Calculate the last digit of each number from A+1 to B for (int i = A + 1; i <= B; ++i) { result *= (i % 10); // Keep only the last digit result %= 10; // Ensure result remains a single digit } return result; } static void Main() { // Example usage Console.WriteLine(LastDigit(2, 4)); // Output: 2 Console.WriteLine(LastDigit(107, 109)); // Output: 2 //This Code is Contributed by Shivam Tiwari }} |
Javascript
function factorial(n) { if (n === 0) { return 1; } return n * factorial(n - 1);}function last_digit(A, B) { var fact_A = factorial(A); var fact_B = factorial(B); return Math.floor(fact_B / fact_A) % 10;}// Example usageconsole.log(last_digit(2, 4)); // Output: 2console.log(last_digit(107, 109)); // Output: 2 |
Output
2 2
Time complexity: O(A+B)
Space complexity: O(A)
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