factorial

I'm studying time complexity in school and our main focus seems to be on polynomial time O(n^c) algorithms and quasi-linear time O(nlog(n)) algorithms with the occasional exponential time O(c^n) algorithm as an example of run-time perspective. However, dealing with larger time complexities was never covered. I would like to see an example problem with an algorithmic solution that runs in factorial time O(n!) . The algorithm may be a naive approach to solve a problem but cannot be artificially bloated to run in factorial time. Extra street-cred if the factorial time algorithm is the best known

So my task is: I have the number 100, and I have to print the sum of the digits of it's factorial. So I wrote the code, found a nice way of summing the digits but my code doesn't work for the number 100. I checked for 10, and it works perfectly. First step that came to my mind, I have to change the type from int to something bigger. I know the result (of the factorial) will be a huge positive number so I choose ulong, but it still doesn't work. I checked around here on Stack Overflow, and the only answers I found suggested using 'BigInteger' but my Visual Studio doesn't seem to know it, and I

Hi this is a factorial method but it prints 0 in the console please help me thanks public class Demo { public static void main(String[] args) { Demo obj = new Demo(); System.out.println(obj.factorial(500)); } public int factorial(int n) { int fact = 1; for (int i = 2; i <= n; i++) { fact= fact*i; } return fact; } EDITED:will return Infinity! public class Demo { public static void main(String[] args) { Demo obj = new Demo(); System.out.println(obj.factorial(500)); } public double factorial(long n) { double fact = 1; for (int i = 2; i <= n; i++) { fact= fact*i; } return fact; } } Since 500!

Having a little trouble trying calculate the number of trailing zeros in a factorial of a given number. This is one of the challenges from Codewars- can't get mine to pass. zeros(12) = 2 #=> 1 * 2 * 3 .. 12 = 479001600 I think I'm on the wrong path here and there is probably a more elegant ruby way. This is what I have down so far. def zeros(n) x = (1..n).reduce(:*).to_s.scan(/[^0]/) return 0 if x == [] return x[-1].length if x != [] end This is more of a math question. And you're right, you are off on a wrong path. (I mean the path you are on is going to lead to a very inefficient solution)

I wrote this function to find a factorial of number fact <- function(n) { if (n < 0){ cat ("Sorry, factorial does not exist for negative numbers", "\n") } else if (n == 0){ cat ("The factorial of 0 is 1", "\n") } else { results = 1 for (i in 1:n){ results = results * i } cat(paste("The factorial of", n ,"is", results, "\n")) } } Now I want to implement Memoization in R. I have Basic idea on R and trying to implement using them. But I am not sure is this way forward. Could you please also elaborate this topic as well. Thank in advance. Memoized Factorial fact_tbl <- c(0, 1, rep(NA, 100)) fact

I have written this function to return the factorial of given number func factorial(_ n: Int) -> Int { if n == 0 { return 1 } else { return n * factorial(n - 1) } } print( factorial(20) ) // 2432902008176640000 Works as it should, as long the given number do not exceed 20, because then the result becomes too high ? How can I circumvent this limit and thus calculate the factorial of higher numbers ? Have searched around and found some bignum libraries for Swift, I'm doing this to learn and be familiar with Swift, therefore I want to figure this out on my own. Here's an approach that will let

Given an array arr of size n , and and index 0<=i<n! I want to return the i'th permutation. I was able to write a method that gets all permutations: function permute (arr) { var permutations = []; if (arr.length === 1) { return [ arr ]; } for (var i = 0; i < arr.length; i++) { var subPerms = permute(arr.slice(0, i).concat(arr.slice(i + 1))); for (var j = 0; j < subPerms.length; j++) { subPerms[j].unshift(arr[i]); permutations.push(subPerms[j]); } } return permutations; } How do I trim it to get only one branch of the recursion ? Nina Scholz You could use the factorial of the array length as