I found an interesting problem on https://www.codewars.com/kata/5550d638a99ddb113e0000a2:
This problem takes its name by arguably the most important event in the life of the ancient historian Josephus: according to his tale, he and his 40 soldiers were trapped in a cave by the Romans during a siege.

Refusing to surrender to the enemy, they instead opted for mass suicide, with a twist: **they formed a circle and proceeded to kill one man every three, until one last man was left (and that it was supposed to kill himself to end the act)**.

Well, Josephus and another man were the last two and, as we now know every detail of the story, you may have correctly guessed that they didn't exactly follow through the original idea.

You are now to create a function that returns a Josephus permutation, taking as parameters the initial *array/list of items* to be permuted as if they were in a circle and counted out every *k* places until none remained.

**Tips and notes:** it helps to start counting from 1 up to n, instead of the usual range 0..n-1; k will always be >=1.

For example, with n=7 and k=3 `josephus(7,3)` should act this way.
```
[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out and goes into the result [3]
[1,2,4,5,7] => 6 is counted out and goes into the result [3,6]
[1,4,5,7] => 2 is counted out and goes into the result [3,6,2]
[1,4,5] => 7 is counted out and goes into the result [3,6,2,7]
[1,4] => 5 is counted out and goes into the result [3,6,2,7,5]
[4] => 1 is counted out and goes into the result [3,6,2,7,5,1]
[] => 4 is counted out and goes into the result [3,6,2,7,5,1,4]
```
So our final result is:
```
josephus([1,2,3,4,5,6,7],3)==[3,6,2,7,5,1,4]
```
For more info, browse the Josephus Permutation page on wikipedia; related kata: Josephus Survivor.

Also, [live game demo](https://iguacel.github.io/josephus/) by [OmniZoetrope](https://www.codewars.com/users/OmniZoetrope).

I tried it in Python, but could not do it. Can you solve it?
def josephus(itemlist, interval):
    length = len(itemlist)

    if interval == 1:
        return itemlist
    
    count = 1
    disposal = [] #list which stores order in which items get deleted

    while len(disposal) != length: #loop runs unti; every item has been deleted
        index = 0 #gives an index. Useful if there are duplicates of items in a list
        for item in itemlist:
            if count == interval: 
                
                disposal.append(item)
                itemlist.pop(index)
                count = 1  
                
                try:
                    print(itemlist[index]) #sees if there is another number after the deleted one. Catches numbers at the end of the list which get's deleted
                except:
                    break

            count += 1
            index += 1
    return disposal