Answer:
O(n)
Explanation:
If we chose to push an element at the end of the list instead of the front, the push operation will have a complexity of O(n) where 'n' represents the number of existing elements of the list. This is because the operation will involve iterating through the entire list of size n via pointers and adding the new element at the end. So the time complexity will be n*O(1) or O(n).
