In this assignment you will practice writing recursion and dynamic programming in a pair of exercises. There is also an optional harder followup to the second exercise.
In this exercise you will
- Practice writing recursive methods
- Practice using dynamic programming techniques
- Determine time & space complexities of recursive methods
Earlier you saw how an array could be used to store Fibonacci numbers resulting in a time-complexity reduction from O(2n) to O(n). Now we will take this a set further, because to find a given Fibonacci number, you only need to find the previous two numbers.
Reminder:
Fib(0) = 0 Fib(1) = 1
Fib(n) = Fib(n-2) + Fib(n-1), for all n >= 2
Restrictions:
- You cannot use a loop, use recursion.
- Your solution should be O(n) space complexity, due to the call stack.
We define super digit of an integer using the following rules:
Given an integer, we need to find the super digit of the integer.
If the number has only digit, then its super digit is that number.
Otherwise, the super digit of x is equal to the super digit of the sum of the digits of x.
For example, the super digit of 9875 will be calculated as:
super_digit(9875) --> superdigit(9 + 8 + 7 + 5) = superdigit(29)
super_digit(29) --> superdigit(2 + 9) = superdigit(11)
super_digit(11) --> superdigit(1 + 1) = superdigit(2)
super_digit(2) --> 2
So the super_digit of 9875 is 2.
In this exercise we will build on the Superdigit concept. A refined superdigit is a determined by concatenating a numbrer n a specific k number of times and then calculating the superdigit.
For example if k = 3 and n = 148
refined_superdigit(n, k) = superdigit(148148148)
= superdigit(1 + 4 + 8 + 1 + 4 + 8 + 1 + 4 + 8)
= superdigit(39)
= superdigit(3 + 9)
= superdigit(12)
= superdigit(1 + 2)
= 3
You can use your superdigit solution in the solution for refined_superdigit. Can you reduce the time complexity?