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.

## Answer

First, let’s look up what the books means with a “Dynamic Array”:

Dynamic array (also called as

growable array,resizable array,

dynamic table, orarray list) is a random access, variable-size list data structure that allows elements to be added or removed.

[…]Note:We will see the implementation fordynamic arrayin 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)`

.