Question

I am trying to move even numbers in an array to the front and odd numbers to the back of the array. The problem asks to do this in a Linear Algorithm and do this In Place.

I came up with this:

 def sort(a):
    for i in range(0,len(a)-1):
        if a[i]%2==0:
            a.insert(0,a.pop(i))
    return a

The issue is that, someone told me that technically, a.insert is an o(n) function so technically this would be considered a non-linear algorithm (when including the for i in range part of the function). Since the forum that asked this question is a couple months old, I couldn’t ask for an explanation.

Basically I believe he said “Technically” because since this inserts it at the front, it does not check another N number of elements in the array, therefore making it run for practical purposes at O(n) and not O(n^2). Is this a correct assessment?

Also, someone on the forum used a.append to modify the above and changed it to look for odd numbers. No one replied so I was wondering, is a.append not an o(n) function since it moves it to the end? Is it o(1)?

Thanks for explanations and clarifications!

Answer 1

insert at the 0th index of a list requires shifting every other element along which makes it an O(N) operation. However, if you use a deque this operation is O(1).

append is an amortized O(1) operation since it simply requires adding the item on to the end of the list and no shifting is done. Sometimes the list needs to grow so it is not always an O(1) operation.