Gfrktyl

I am currently reading my textbook and I am totally confused why a dynamic array would require O(n) time to delete an item at the end. I understand that deleting an item from any other index is O(n) because you have to copy all the data and move them to fill in the gap, but if it’s at the end don’t we simply just decrement the count and set the index to like 0 or null? I included a picture from my book. It’s weird cause it says indexing is O(1) so we must know where the item is so we don’t have to traverse the array like a linked list.

First, let's look up what the books means with a "Dynamic Array":

Dynamic array (also called as growable array, resizable array,
dynamic table, or array list) is a random access, variable-size list data structure that allows elements to be added or removed.
[...]
Note: We will see the implementation for dynamic array in the Stacks, Queues and Hashing chapters.

From this we learn that array lists are examples of a "Dynamic Array" as the author of the book defines it.

But looking further, the book mentioned that:

As soon as that array becomes full, create the new array of size
double than the original array. Similarly, reduce the array size to
half if the elements in the array are less than half.

(emphasis added by me)

A Java ArrayList doesn't do this - it doesn't decrease in storage when elements are removed. But the author is talking about (or believes that ArrayList does) reduce the array size.
In that case, from a worst-worst-case perspective, you could say that the complexity is O(n) because reducing the size involves copying n elements to the reduced array.

Conclusion:

Although it's not true for Java ArrayList implementations, when the author of this book talks about "dynamic arrays" that "reduce the array size" on deletion when necessary, then the worst-case complexity of a delete at the end of the array is indeed O(n).

That entry seems like it's either

1. incorrect, or


2. true, but misleading.

You are absolutely right that you can just destroy the object at the final position in a dynamic array and then decrement the size to remove the last element. In many implementations of dynamic arrays, you'll sometimes need to perform resize operations to make sure that the size of the allocated array is within some constant factor of the number of elements. If that happens, then yes, you'll need to make a new array, copy over the old elements, and free the previous array, which does take time O(n). However, you can show that these resizes are sufficiently infrequent that the average cost of doing a remove from the end is O(1). (In a more technical sense, we say that the amortized cost of removing an element from the end is O(1)). That is, as long as you care only about the total cost of performing a series of operations rather than the cost of any individual operation, you would not be wrong to just pretend each operation costs you time O(1).

I'd say that you're most likely looking at a typo in the materials. Looking at their entry for appending to the end, which differentiates between the not-full and full cases, I think this is likely a copy/paste error. Following the table's lead, that should say something to the effect of "O(1) if the array is not 'too empty,' O(n) otherwise." But again, keep in mind that the amortized efficiency of each operation is O(1), meaning that these scary O(n) terms aren't actually likely to burn you in practice unless you are in a specialized environment where each operation needs to work really quickly.

In java for Dynamic array (ArrayList) time complexity deletion of last element is o(1) in java it does not copy array
in java they will check weather the array index is end.

  int numMoved = size - index - 1;

        if (numMoved > 0)

         //copy array element 

Insertion and deletion are operations that we generally do not perform on arrays, because they have a fixed length by nature. You cannot increase or decrease the length of something which is fixed by its nature.

When people speak of "dynamic arrays" in Java they tend to mean using class ArrayList, which is backed by an array, and it provides the illusion of the ability to insert and remove elements.

But in order for some piece of software to provide the illusion of performing insertions and deletions on an array, each time (or almost each time, there are optimizations possible) it has to allocate a new array of the new desired length, and copy the old array into it, skipping the removed element or adding the inserted element as the case may be. That array copy is where the O(N) comes from.

And, due to the optimizations performed by ArrayList, the table that you posted is not accurate: it should rather say 'O(1) if the array has not shrunk by much, O(N) if the array has shrunk by so much that reallocation is deemed necessary'. But I guess that would have been too long to fit in the table cell.

As you have mentioned this could be confusing, if you add an element to a dynamic array, it change its size in a constant interval and will create a new array an copy elements to the new array as you may already know. And when it shrinks in size it will also shirk if needed.

For an example if interval is 4 when you add 1st, 2nd, 3rd, 4th element, everything will be okay, but when you add the 5th item dynamic array will grow into a 8 elements array and will copy the all elements to the new array.

It is the same when it is decreasing. If you remove one item from a 5 item array which has an interval of 4, dynamic array it will create a new 4 elements array and copy the elements.

Here is a good representation video tutorial,

Yes. When dynamic array does not have to shrink it is O(1) which it takes to remove the element but when it has to shrink its O(n), as you may already figured out.

when you find the big O notation you are defining the worst case, so it is O(n)

As far as I think
As it is a dynamic array so the computer system does not know as to what is the current length of this dynamic array so to find the length of this dynamic array it takes O(n) time and then takes O(1) time to delete the element at end.