Smallest number with at least n digits in factorial

Given a number n. The task is to find the smallest number whose factorial contains at least n digits.

Examples:

Input : n = 1
Output : 0
0! = 1, hence it has 1 digit.

Input : n = 2
Output : 4
4! = 24 and 3! = 6, hence 4 is 
the smallest number having 2 
digits in its factorial

Input : n = 5
Output : 8

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

In the article for Count digits in a factorial of a number, we have discussed how we can efficiently find the number of digits in factorial.

We used the below formula to find the number of digits

Kamenetsky’s formula approximates the number
of digits in a factorial by :
f(x) = log₁₀(((n/e)ⁿ) * sqrt(2*pi*n))

Thus, we can pretty easily use the property of logarithms to ,
f(x) = n*log₁₀((n/e)) + log10(2*pi*n)/2

Now we need to determine the interval in which we can find a factorial which has at least n digits. Following are some observations:

For a large number we can always say that it’s factorial has more digits than the number itself. For example factorial of 100 has 158 digits which is greater than 100.
However for smaller numbers, that might not be the case. For example factorial of 8 has only 5 digits, which is less than 8. In fact numbers up to 21 follow this trend.

Hence if we search from 0! to n! to find a result having at least n digits, we won’t be able to find the result for smaller numbers.

For example suppose n = 5, now as we’re searching in [0,n] the maximum number of digits we can obtain is 3, (found in 5! = 120). However if we search in [0, 2*n] (0 to 10), we can find 8! has 5 digits.

Hence, if we can search for all factorial from 0 to 2*n , there will always be a number k which will have at least n digits in its factorial.(Readers are advised to try on their own to figure out this fact)

We can say conclude if we have to find a number k,
such that k! has at least n digits, we can be sure
that k lies in [0,2*n]

i.e., 0<= k <= 2*n

Thus we can do a binary search between 0 to 2*n to find the smallest number having at least n digits.

C++

// A C++ program to find the smallest number 
// having at least n digits in factorial 
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns the number of digits present in n! 
int findDigitsInFactorial(int n) 
{ 
    // factorial of -ve number doesn't exists 
    if (n < 0) 
        return 0; 
  
    // base case 
    if (n <= 1) 
        return 1; 
  
    // Use Kamenetsky formula to calculate the 
    // number of digits 
    double x = ((n*log10(n/M_E)+log10(2*M_PI*n)/2.0)); 
  
    return floor(x)+1; 
} 
  
// This function receives an integer n and returns 
// an integer whose factorial has at least n digits 
int findNum(int n) 
{ 
    // (2*n)! always has more digits than n 
    int low = 0, hi = 2*n; 
  
    // n <= 0 
    if (n <= 0) 
        return -1; 
  
    // case for n = 1 
    if (findDigitsInFactorial(low) == n) 
        return low; 
  
    // now use binary search to find the number 
    while (low <= hi) 
    { 
        int mid = (low+hi) / 2; 
  
        // if (mid-1)! has lesser digits than n 
        // and mid has n or more then mid is the 
        // required number 
        if (findDigits(mid) >= n && findDigits(mid-1)<n) 
            return mid; 
  
        else if (findDigits(mid) < n) 
            low = mid+1; 
  
        else
            hi = mid-1; 
    } 
    return low; 
} 
  
// Driver program to check the above functions 
int main() 
{ 
    cout << findNum(1) << endl; 
    cout << findNum(2) << endl; 
    cout << findNum(5) << endl; 
    cout << findNum(24) << endl; 
    cout << findNum(100) << endl; 
    cout << findNum(1221) << endl; 
  
    return 0; 
} 

Java

// A Java program to find the  
// smallest number having at 
// least n digits in factorial 
class GFG 
{ 
// Returns the number of 
// digits present in n! 
static int findDigitsInFactorial(int n) 
{ 
    // factorial of -ve number  
    // doesn't exists 
    if (n < 0) 
        return 0; 
  
    // base case 
    if (n <= 1) 
        return 1; 
  
    // Use Kamenetsky formula to  
    // calculate the number of digits 
    double x = ((n * Math.log10(n / Math.E) +  
                     Math.log10(2 * Math.PI * n) / 2.0)); 
  
    return (int)(Math.floor(x) + 1); 
} 
  
// This function receives an integer  
// n and returns an integer whose  
// factorial has at least n digits 
static int findNum(int n) 
{ 
    // (2*n)! always has 
    // more digits than n 
    int low = 0, hi = 2 * n; 
  
    // n <= 0 
    if (n <= 0) 
        return -1; 
  
    // case for n = 1 
    if (findDigitsInFactorial(low) == n) 
        return low; 
  
    // now use binary search  
    // to find the number 
    while (low <= hi) 
    { 
        int mid = (low + hi) / 2; 
  
        // if (mid-1)! has lesser digits 
        // than n and mid has n or more  
        // then mid is the required number 
        if (findDigitsInFactorial(mid) >= n && 
            findDigitsInFactorial(mid - 1) < n) 
            return mid; 
  
        else if (findDigitsInFactorial(mid) < n) 
            low = mid + 1; 
  
        else
            hi = mid - 1; 
    } 
    return low; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    System.out.println(findNum(1)); 
    System.out.println(findNum(2)); 
    System.out.println(findNum(5)); 
    System.out.println(findNum(24)); 
    System.out.println(findNum(100)); 
    System.out.println(findNum(1221)); 
} 
} 
  
// This Code is Contributed by mits 

Python3

# Python3 program to find the  
# smallest number 
# having at least n digits  
# in factorial 
  
import math 
  
# Returns the number of digits  
# present in n! 
def findDigitsInFactorial(n): 
      
    # factorial of -ve number  
    # doesn't exists 
    if (n < 0): 
        return 0
  
    # base case 
    if (n <= 1): 
        return 1
  
    # Use Kamenetsky formula to calculate the 
    # number of digits 
    M_E=2.7182818284590452354
    M_PI=3.14159265358979323846
    x = ((n*math.log10(n/M_E)+math.log10(2*M_PI*n)/2.0)) 
  
    return int(math.floor(x)+1) 
  
# This function receives an  
# integer n and returns 
# an integer whose factorial has  
# at least n digits 
def findNum(n): 
      
    # (2*n)! always has more  
    # digits than n 
    low = 0
    hi = 2*n 
  
    # n <= 0 
    if (n <= 0): 
        return -1
  
    # case for n = 1 
    if (findDigitsInFactorial(low) == n): 
        return int(round(low)) 
  
    # now use binary search to  
    # find the number 
    while (low <= hi): 
        mid = int((low+hi) / 2) 
  
        # if (mid-1)! has lesser digits than n 
        # and mid has n or more then mid is the 
        # required number 
        if ((findDigitsInFactorial(mid) >= n and 
            findDigitsInFactorial(mid-1)<n)): 
            return int(round(mid)) 
  
        elif (findDigitsInFactorial(mid) < n): 
            low = mid+1
  
        else: 
            hi = mid-1
              
    return int(round(low)) 
  
# Driver code 
if __name__=='__main__': 
    print(findNum(1)); 
    print(findNum(2)); 
    print(findNum(5)); 
    print(findNum(24)); 
    print(findNum(100)); 
    print(findNum(1221)); 
  
# this code is contributed by  
# mits 

C#

// A C# program to find the  
// smallest number having at  
// least n digits in factorial  
using System; 
  
class GFG 
{ 
      
// Returns the number of  
// digits present in n!  
static int findDigitsInFactorial(int n)  
{  
    // factorial of -ve number  
    // doesn't exists  
    if (n < 0)  
        return 0;  
  
    // base case  
    if (n <= 1)  
        return 1;  
  
    // Use Kamenetsky formula to  
    // calculate the number of digits  
    double x = ((n * Math.Log10(n / Math.E) +  
                     Math.Log10(2 * Math.PI * n) / 2.0));  
  
    return (int)(Math.Floor(x) + 1);  
}  
  
// This function receives an integer  
// n and returns an integer whose  
// factorial has at least n digits  
static int findNum(int n)  
{  
    // (2*n)! always has  
    // more digits than n  
    int low = 0, hi = 2 * n;  
  
    // n <= 0  
    if (n <= 0)  
        return -1;  
  
    // case for n = 1  
    if (findDigitsInFactorial(low) == n)  
        return low;  
  
    // now use binary search  
    // to find the number  
    while (low <= hi)  
    {  
        int mid = (low + hi) / 2;  
  
        // if (mid-1)! has lesser digits  
        // than n and mid has n or more  
        // then mid is the required number  
        if (findDigitsInFactorial(mid) >= n &&  
            findDigitsInFactorial(mid - 1) < n)  
            return mid;  
  
        else if (findDigitsInFactorial(mid) < n)  
            low = mid + 1;  
  
        else
            hi = mid - 1;  
    }  
    return low;  
}  
  
// Driver Code  
static public void Main () 
{ 
    Console.WriteLine(findNum(1));  
    Console.WriteLine(findNum(2));  
    Console.WriteLine(findNum(5));  
    Console.WriteLine(findNum(24));  
    Console.WriteLine(findNum(100));  
    Console.WriteLine(findNum(1221));  
}  
}  
  
// This code is contributed by akt_mit  

PHP

<?php 
// A PHP program to find the smallest number 
// having at least n digits in factorial 
  
// Returns the number of digits  
// present in n! 
function findDigitsInFactorial($n) 
{ 
    // factorial of -ve number  
    // doesn't exists 
    if ($n < 0) 
        return 0; 
  
    // base case 
    if ($n <= 1) 
        return 1; 
  
    // Use Kamenetsky formula to  
    // calculate the number of digits 
    $x = (($n * log10($n / M_E) +  
                log10(2 * M_PI * $n) / 2.0)); 
  
    return (int)(floor($x) + 1); 
} 
  
// This function receives an integer  
// n and returns an integer whose  
// factorial has at least n digits 
function findNum($n) 
{ 
    // (2*n)! always has more  
    // digits than n 
    $low = 0; 
    $hi = 2 * $n; 
  
    // n <= 0 
    if ($n <= 0) 
        return -1; 
  
    // case for n = 1 
    if (findDigitsInFactorial($low) == $n) 
        return (int)round($low); 
  
    // now use binary search to  
    // find the number 
    while ($low <= $hi) 
    { 
        $mid = ($low + $hi) / 2; 
  
        // if (mid-1)! has lesser digits  
        // than n and mid has n or more  
        // then mid is the required number 
        if (findDigitsInFactorial($mid) >= $n && 
            findDigitsInFactorial($mid - 1) < $n) 
            return (int)round($mid); 
  
        else if (findDigitsInFactorial($mid) < $n) 
            $low = $mid + 1; 
  
        else
            $hi = $mid - 1; 
    } 
    return (int)round($low); 
} 
  
// Driver Code 
echo findNum(1) . "\n"; 
echo findNum(2) . "\n"; 
echo findNum(5) . "\n"; 
echo findNum(24) . "\n"; 
echo findNum(100) . "\n"; 
echo findNum(1221) . "\n"; 
  
// This code is contributed by mits 
?> 

Output:

Complexity Analysis
The complexity for the binary search is O(log(2*n)), if we ignore the complexity of the logarithmic function. Hence overall complexity is O(log(n)).

This article is contributed by Ashutosh Kumar If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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