Count digits in a factorial | Set 1
Given an integer n, find the number of digits that appear in its factorial, where factorial is defined as, factorial(n) = 1*2*3*4……..*n and factorial(0) = 1
Examples :
Input : n = 1 Output : 1 1! = 1 , hence number of digits is 1 Input : 5 Output : 3 5! = 120, i.e., 3 digits Input : 10 Output : 7 10! = 3628800, i.e., 7 digits
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A naive solution would be to calculate the n! first and then calculate the number of digits present in it. However as the value for n! can be very large, it would become cumbersome to store them in a variable (Unless you’re working in python!) .
A better solution would be to use the useful property of logarithms to calculate the required answer.
We know,
log(a*b) = log(a) + log(b)
Therefore
log( n! ) = log(1*2*3....... * n)
= log(1) + log(2) + ........ +log(n)
Now, observe that the floor value of log base
10 increased by 1, of any number, gives the
number of digits present in that number.
Hence, output would be : floor(log(n!)) + 1.Below is a the implementation of the same.
C++
// A C++ program to find the number of digits in// a factorial#include <bits/stdc++.h>using namespace std;// This function receives an integer n, and returns// the number of digits present in n!int findDigits(int n){ // factorial exists only for n>=0 if (n < 0) return 0; // base case if (n <= 1) return 1; // else iterate through n and calculate the // value double digits = 0; for (int i=2; i<=n; i++) digits += log10(i); return floor(digits) + 1;}// Driver codeint main(){ cout << findDigits(1) << endl; cout << findDigits(5) << endl; cout << findDigits(10) << endl; cout << findDigits(120) << endl; return 0;} |
Java
// Java program to find the number// of digits in a factorialimport java.io.*;import java.util.*;class GFG{ // returns the number of digits // present in n! static int findDigits(int n) { // factorial exists only for n>=0 if (n < 0) return 0; // base case if (n <= 1) return 1; // else iterate through n and calculate the // value double digits = 0; for (int i=2; i<=n; i++) digits += Math.log10(i); return (int)(Math.floor(digits)) + 1; } // Driver code public static void main (String[] args) { System.out.println(findDigits(1)); System.out.println(findDigits(5)); System.out.println(findDigits(10)); System.out.println(findDigits(120)); }}// This code is contributed by Pramod Kumar |
Python3
# Python3 program to find the# number of digits in a factorialimport math# This function receives an integer# n, and returns the number of# digits present in n!def findDigits(n): # factorial exists only for n>=0 if (n < 0): return 0; # base case if (n <= 1): return 1; # else iterate through n and # calculate the value digits = 0; for i in range(2, n + 1): digits += math.log10(i); return math.floor(digits) + 1;# Driver codeprint(findDigits(1));print(findDigits(5));print(findDigits(10));print(findDigits(120));# This code is contributed by mits |
C#
// A C++ program to find the number// of digits in a factorialusing System;class GFG { // This function receives an integer // n, and returns the number of // digits present in n! static int findDigits(int n) { // factorial exists only for n>=0 if (n < 0) return 0; // base case if (n <= 1) return 1; // else iterate through n and // calculate the value double digits = 0; for (int i = 2; i <= n; i++) digits += Math.Log10(i); return (int)Math.Floor(digits) + 1; } // Driver code public static void Main() { Console.Write(findDigits(1) + "\n"); Console.Write(findDigits(5) + "\n"); Console.Write(findDigits(10) + "\n"); Console.Write(findDigits(120) + "\n"); }}// This code is contributed by// Smitha Dinesh Semwal |
PHP
<?php// PHP program to find// the number of digits// in a factorial// This function receives// an integer n, and returns// the number of digits present in n!function findDigits($n){ // factorial exists only for n>=0 if ($n < 0) return 0; // base case if ($n <= 1) return 1; // else iterate through n and // calculate the value $digits = 0; for ($i = 2; $i <= $n; $i++) $digits += log10($i); return floor($digits) + 1;}// Driver codeecho findDigits(1), "\n";echo findDigits(5), "\n";echo findDigits(10), "\n";echo findDigits(120), "\n";// This code is contributed by Ajit.?> |
Javascript
<script>// A Javascript program to find the number of digits in// a factorial// This function receives an integer n, and returns// the number of digits present in n!function findDigits(n){ // factorial exists only for n>=0 if (n < 0) return 0; // base case if (n <= 1) return 1; // else iterate through n and calculate the // value let digits = 0; for (let i=2; i<=n; i++) digits += Math.log10(i); return Math.floor(digits) + 1;}// Driver code document.write(findDigits(1) + "<br>"); document.write(findDigits(5) + "<br>"); document.write(findDigits(10) + "<br>"); document.write(findDigits(120) + "<br>"); //This code is contributed by Mayank Tyagi</script> |
Output :
1 3 7 199
In the next set, we’d see how to further optimize our approach and reduce the time complexity for the same program.
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