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# 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```

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 ``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)

## 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``)

## 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

 `= ``\$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``?>`

## Javascript

 ``

Output:

```0
4
8
24
70
532```

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)).

Space complexity: O(1) since using constant space

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