Given an integer n (can be very large), 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
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 Input : 50000000 Output : 363233781 Input : 1000000000 Output : 8565705523
We’ve already discussed the solution for small values of n in the Count digits in a factorial | Set 1. However that solution would not be able to handle cases where n >10^6
So, can we improve our solution ?
Yes ! we can.
We can use Kamenetsky’s formula to find our answer !
It 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
And that’s it !
Our solution can handle very large inputs that can be accommodated in a 32 bit integer,
and even beyond that ! .
Below is the implementation of above idea :
1 363233781 8565705523 199
References : oeis.org
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- Count digits in a factorial | Set 1
- Smallest number with at least n digits in factorial
- Find sum of digits in factorial of a number
- Number of digits in N factorial to the power N
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count Divisors of Factorial
- Count numbers in given range such that sum of even digits is greater than sum of odd digits
- Count factorial numbers in a given range
- Count trailing zeroes in factorial of a number
- Find the last digit when factorial of A divides factorial of B
- Count of integers in a range which have even number of odd digits and odd number of even digits
- Count even and odd digits in an Integer
- Count numbers with same first and last digits
- Count of integers of length N and value less than K such that they contain digits only from the given set
- Count total number of digits from 1 to n