Q: Iterative version of this recursive function I have a function that is defined in this way:

F(n) = n if n<=3
F(n) = F(n-1) + 2 * F(n-2) + 3 * F(n-3) if n>3

Now, I've written it as a recursive function and it works fine.
I'm trying to write it as an iterative function but i cant seem to make it happen.
The output should be, for example:
print(FRec(5))  =>     22
print(FRec(10)) =>   1657
print(FRec(15)) => 124905

Any tips?
Here is my recursive implementation:
def FRec(n):
    if(n <= 3):
        return n
    if(n > 3):
        return FRec(n - 1) + 2 * FRec(n - 2) + 3 * FRec(n - 3)


A: All you need is keep the last 3 results:
from collections import deque

def F_iter(n):
    if n <= 3:
        return n
    prev = deque([1, 2, 3], maxlen=3)
    for i in range(4, n + 1):
        result = prev[-1] + 2 * prev[-2] + 3 * prev[-3]
        prev.append(result)
    return prev[-1]

If a deque is not 'available' to you, then you can inefficiently replace that with some list slicing and concatenation:
    prev = [1, 2, 3]
    for i in range(4, n + 1):
        result = prev[-1] + 2 * prev[-2] + 3 * prev[-3]
        prev = prev[1:] + [result]

Demo:
>>> F_iter(5)
22
>>> F_iter(10)
1657
>>> F_iter(15)
124905

