Gain insights into array-based and linked list-based data structures, explore advanced structures like skiplists and hashing, and discover template-based collections for efficient data storage and retrieval.

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Data structures and algorithms are essential in computer science since they play a crucial role in efficient information retrieval and processing, dealing with files, storing contacts on phones, social networks and web searches.

In this course, you’ll learn about the array-based implementation of various linear data structures, stack, and queues. You’ll also learn about linked list-based implementation. Next, you’ll explore advanced data structures like skiplists and hashing. You’ll learn how to implement a variety of trees and graphs, and data structures related to bits of an integer. Toward the end of the course, you’ll learn the implementation of structures based on external storage.

After completing this course, you’ll be able to create reusable programs with template-based collections that can efficiently analyze how to optimize the storage and retrieval of very large amounts of data. Overall, this course will enhance your productivity and performance as a software developer.

Data Structures with Generic Types in Python

# Growing and shrinking
Note that, unlike the `ArrayStack.resize()` operation, `grow()` and `shrink()` do not copy any data. They only allocate or free an array of size $r$. In some environments, this takes only constant time, while in others, it may require time proportional to $r$.

We note that, immediately after a call to `grow()` or `shrink()`, the situation is clear. The final block is completely empty, and all other blocks are completely full. Another call to `grow()` or `shrink()` will not happen until at least $r − 1$ elements have been added or removed. Therefore, even if `grow()` and `shrink()` take $O(r)$ time, this cost can be amortized over at least $r − 1$ `add(i, x)` or `remove(i)` operations, so that the amortized cost of `grow()` and `shrink()` is $O(1)$ per operation.

# Space usage
Next, we analyze the amount of extra space used by a `RootishArrayStack`. In particular, we want to count any space used by a `RootishArrayStack` that is not an array element currently used to hold a list element. We call all such space **wasted space**.

The `remove(i)` operation ensures that a `RootishArrayStack` never has more than two blocks that are not completely full. The number of blocks, $r$, used by a `RootishArrayStack` that stores $n$ elements therefore satisfies

$$
(r − 2)(r − 1) \le n 
$$

Again, using the quadratic equation on this gives

$$ 
r \le (3 + \sqrt{1+4n})/2 = O(\sqrt n)
$$

The last two blocks have sizes $r$ and $r − 1$, so the space wasted by these two blocks is at most $2r−1 = O(\sqrt n)$. If we store the blocks in (for example) an `ArrayStack`, then the amount of space wasted by the List that stores those $r$ blocks is also $O(r) = O(\sqrt n)$. The other space needed for storing $n$ and other accounting information is $O(1)$. Therefore, the total amount of wasted space in a `RootishArrayStack` is $O(\sqrt n)$.

Next, we argue that this space usage is optimal for any data structure that starts out empty and can support the addition of one item at a time. More precisely, we will show that, at some point during the addition of n items, the data structure is wasting an amount of space at least in $\sqrt n$ (though it may be only wasted for a moment).

Suppose we start with an empty data structure and we add $n$ items one at a time. At the end of this process, all $n$ items are stored in the structure and distributed among a collection of $r$ memory blocks. If $r \ge \sqrt n$, then the data structure must be using $r$ pointers (or references) to keep track of these $r$ blocks, and these pointers are wasted space. On the other hand, if $r &lt; \sqrt n$ then, by the pigeonhole principle, some block must have a size of at least $n/r &gt; \sqrt n$. Consider the moment at which this block was first allocated. Immediately after it was allocated, this block was empty, and was therefore wasting $\sqrt n$ space. Therefore, at some point in time during the insertion of $n$ elements, the data structure was wasting $\sqrt n$ space.

# Analysis of `RootishArrayStack`
The following theorem summarizes our discussion of the `RootishArrayStack` data structure:

**Theorem:** A `RootishArrayStack` implements the List interface. Ignoring the cost of calls to `grow()` and `shrink()`, a `RootishArrayStack` supports the operations

- `get(i)` and `set(i, x)` in $O(1)$ time per operation; and
- `add(i, x)` and `remove(i)` in $O(1+n−i)$ time per operation.

Furthermore, beginning with an empty `RootishArrayStack`, any sequence of $m$ `add(i, x)` and `remove(i)` operations results in a total of $O(m)$ time spent during all calls to `grow()` and `shrink()`.

The space measured in words used by a `RootishArrayStack` that stores $n$ elements is $n + O(\sqrt n)$.


Explore the advantage of using RootishArrayStack.

Overview

Array-Based Lists

Linked Lists

Skiplists

Hash Tables

Binary Trees

Random Binary Search Trees

Scapegoat Trees

Red-Black Trees

Heaps

Sorting Algorithms

Graphs

Data Structures for Integers

External Memory Searching

Wrap Up

Analysis of RootishArrayStack

Growing and shrinking