Rearrange the below algorithm for computing n Factorial.Input: n, an integer greater than or equal to 0Output: n!end factorialreturn(1)return(n * f actorial(n - 1))procedure factorial( n)if n = 0 then
Question
Rearrange the below algorithm for computing n Factorial.Input: n, an integer greater than or equal to 0Output: n!end factorialreturn(1)return(n * f actorial(n - 1))procedure factorial( n)if n = 0 then
Solution
Sure, here is the rearranged algorithm for computing n Factorial:
procedure factorial(n)
if n = 0 then
return(1)
else
return(n * factorial(n - 1))
end factorial
This algorithm starts by defining a procedure called "factorial" that takes an integer "n" as input. If "n" is 0, it returns 1 (since the factorial of 0 is 1). If "n" is not 0, it returns the product of "n" and the factorial of "n-1". This is a recursive call, meaning the function calls itself with a smaller input. The recursion continues until it reaches the base case where n is 0.
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