# Limitation of arrays

Although arrays are useful data structures, they have limitations:

• the size of the array is fixed at compile time.
• When overflow happens, a copy of the old array is needed.
• the data elements are physically arranged in memory next to each other. This has both advantages and limitations.

## How to compare different data structures?

We already studied array lists and arrays. How do we compare the performance of these data-structures, especially as we perform different operations such as inserting, deleting at different positions of the datastructure. To perform such comparisons without the need for actual implementations, the Big-O notation is used. Let’s see a few examples to illustrate the concept:

## An informal introduction to Big-O

Thus O expresses an upper-bound on the total number of operations performed by some function or algorithm as a function of the number of inputs N.

• O(1) – a constant number of operations, that is independent of the input size.
• O(N) – the number of operations is a polynomial whose highest degree is N (linear).
• O($$N^2$$) – the number of operations is a polynomial whose highest degree is $$N^2$$ (quadratic), and so on.

if-statement, where the body is independent of the input size: O(1)

if (a < b) {
tmp = a;
a = b;
b = tmp;
}


for-loop, while-loop, with linear steps: O(N)

for (i = 0; i < N; i += 2)


for-loop, while-loop, with dividing/multiplying steps: O(logN)

for (i = 1; i < N; i*=2)


nested for-loop, linear steps: O($$N^2$$)

for (i = 0; i < N; i+=2)
for (j = 0; j < N; j++)


for-loop with function call… in this example: O($$N$$logN)

for (i = 0; i < N; i+=2)  // O(N)
divide();      // O(logN)


nested for-loop with function call… in this example: O($$N^3$$logN)

for (i = 0; i < N; i++)           // O(N)
for (j = 1; j < N; j*=2)      // O(logN)
bubbleSort(list);  // O(N*N)


## Array complexity

• Adding a new element: O(N), copying all elements to a larger array

• Deleting an element: O(N) if in middle/front, O(1) at the end

• Retrieving an element at a particular index: O(1)

## Array list complexity

• Adding new value: O(1) amortised (on average) if doubling size… N insertions, and doubling array k times, to fit N elements (so N $$\approx {2^k}$$)
• Total cost: $$N + 2^1 + 2^2 + 2^3 + ... + 2^{k-1} + 2^k \\\rightarrow N + 2\times2^k (since \sum(2^1 + 2^2 + 2^3 + ... + 2^{k-1}) \approx 2^k) \\\rightarrow N + 2N \\\rightarrow O(N)$$
• The total cost is O(N), amortized to O(1) over N operations

• Add/delete at front: O(1)

• Add/delete at end, without knowing tail: ?

• Add at end, with knowing tail: ?

• Delete at end, with knowing tail, without access to prev node: ?

• Delete at end, with knowing tail, with access to prev node: ?

• Add/delete in middle: ?

• Retrieving an element at a particular index: ?

# Linked List building block – a Node

• A Node is an object which consists of a value and a reference to the next Node.
• Here is a potential implementation:

public class Node {
private int val;
private Node next;

// constructor
public Node(int v) {
val = v;
next = null;
}

// getters and setters

public void setNext(Node n) {
next = n;
}

public Node getNext() {
return next;
}

public int getValue() {
return val;
}

public static void main(String args[]) {
Node n1 = new Node(0);
Node n2 = new Node(200);
Node n3 = new Node(-50);
n1.setNext(n2);
n2.setNext(n3);
n3.setNext(null);

Node n = new Node();
n = n1;
while (n != null) {
System.out.println("The value of the next node is: " + n.getValue());
n = n.getNext();
}
}
}


## What are the limitations of the above Node?

• The Node stores integer values only. What we need to store other types?
• There are problems with the main method as well. We will defer this discussion to later.

However, isn’t this problem easy to fix, since Java supports the concept of Objects. Here is a revision of the previous implementation.

public class Node {
private Object val;
private Node next;

// constructor

public Node() {

}

public Node(Object v) {
val = v;
next = null;
}

// getters and setters

public void setNext(Node n) {
next = n;
}

public Node getNext() {
return next;
}

public Object getValue() {
return val;
}

public static void main(String args[]) {
Node n1 = new Node(0);
Node n2 = new Node(200);
Node n3 = new Node(-50);
n1.setNext(n2);
n2.setNext(n3);
n3.setNext(null);

Node n = new Node();
n = n1;
Double sum = 0.0;
while (n != null) {
System.out.println("The value of the next node is: " + n.getValue());
sum = sum + ((Double) n.getValue());
n = n.getNext();
}
System.out.println("The sum of the values: " + sum);
}
}


What problem do you find with this approach?

# Generics

• The key idea is to support parametric types
• Thus, classes, interfaces operate on data-types, which can be changed like a parameter.
• Classes, interfaces and methods operating on parametric types are called generic classes, interfaces or methods.
• Allows us to define a data-structure or algorithm once, which is agnostic to the data-types it will operate on.
• Ensure type-safe operation

## Terminology used for Java Types ## Type safety with Java Generics

• For a review of Java data types examine the following link
• One of the main benefits of generics is to ensure type safety
• Type safety ensures that expressions and assignments are done using consistent types
• This ensures that what the programmer intends at compile time happens at run time.
• Here is how the previous program is revised using Java generics to ensure type safety
• NOTE: In the following T can’t be a primitive type but has to be a reference type i.e. you can use any object except built in data types such as int, float etc.

public class Node<T> {
private T val;
private Node<T> next;

// constructor

public Node() {

}

public Node(T v) {
val = v;
next = null;
}

// getters and setters

public void setNext(Node<T> n) {
next = n;
}

public Node<T> getNext() {
return next;
}

public T getValue() {
return val;
}

public static void main(String args[]) {
Node<Integer> n1 = new Node<Integer>(0);
Node<Integer> n2 = new Node<Integer>(200);
Node<Integer> n3 = new Node<Integer>(-50);
n1.setNext(n2);
n2.setNext(n3);
n3.setNext(null);

Node<Integer> n = new Node<Integer>();
n = n1;
Double sum = 0.0;
while (n != null) {
System.out.println("The value of the next node is: " + n.getValue());
sum += n.getValue();
n = n.getNext();
}

System.out.println("The sum of the values: " + sum);
}
}


## How about adding node values of different types

public class Node<T> {
private T val;
private Node<T> next;

// constructor

public Node() {

}

public Node(T v) {
val = v;
next = null;
}

// getters and setters

public void setNext(Node<T> n) {
next = n;
}

public Node<T> getNext() {
return next;
}

public T getValue() {
return val;
}

public static void main(String args[]) {
Node<String> n1 = new Node<String>("se");
Node<String> n2 = new Node<String>("2");
Node<String> n3 = new Node<String>("8");
Node<String> n4 = new Node<String>("1 ");

n1.setNext(n2);
n2.setNext(n3);
n3.setNext(n4);
n4.setNext(null);

Node<String> n = new Node<String>();
n = n1;
String sum = "";
while (n != null) {
System.out.println("The value of the next node is: " + n.getValue());

sum = sum + n.getValue();
n = n.getNext();
}

System.out.println("The sum of the values: " + sum);
sum = sum + new Node<Object>(n1).getValue();
System.out.println("The next sum of the values: " + sum);
}
}


## Bounded Types

public class Node<T extends Number> {
private T val;
private Node<T> next;

// constructor

public Node() {

}

public Node(T v) {
val = v;
next = null;
}

// getters and setters

public void setNext(Node<T> n) {
next = n;
}

public Node<T> getNext() {
return next;
}

public T getValue() {
return val;
}

public static void main(String args[]) {
Node<Integer> n1 = new Node<Integer>(0);
Node<Integer> n2 = new Node<Integer>(200);
Node<Integer> n3 = new Node<Integer>(-50);
n1.setNext(n2);
n2.setNext(n3);
n3.setNext(null);

Node<Integer> n = new Node<Integer>();
n = n1;
Double sum = 0.0;
while (n != null) {
System.out.println("The value of the next node is: " + n.getValue());

sum = sum + n.getValue();
n = n.getNext();
}

System.out.println("The sum of the values: " + sum);

sum += new Node<Double>(10.0).getValue();

System.out.println("The sum of the values: " + sum);

}
}


## Linked List by linking nodes nodes (see above)

This approach is not good OO design. Here are some points why?

• A Node is part of a LinkedList but is not the list itself
• The API of the Node class deals with method for a Node rather than for a list
• A LinkedList may be of many types, i.e. a singly linked list or a doubly linked list. Here we can use the familiar interfaces to ensure that both types have the same interface.

public interface List<T> {
public void append(T item);

public void prepend(T item);

public T fetch(int pos) throws InvalidPositionException;

public void insert(int pos, T data) throws InvalidPositionException;

public void remove(int pos) throws InvalidPositionException;

public int size();

public void print();

}


# Get your hands dirty!

In the following you are given the first version of the LinkedList that implements the List interface.

public class LinkedList<T> implements List<T> {

}

// Key methods of the List interface

public void prepend(T data) {
// Note -- works even if list is empty
Node<T> n = new Node<T>(data);
}

public void append(T data) {

}

public T fetch(int pos) throws InvalidPositionException {
T val = head.getValue();
return val;
}

public void insert(int pos, T data) throws InvalidPositionException {

}

public void remove(int pos) throws InvalidPositionException {

}

public int size() {
int i = 0;
return i;
}

}



We will now do the following exercise in ACP:

1. In ACP you are already given the test case for testing the prepend method for prepending integers. This fails. Why?
2. Implement the appropriate fix so that the test on the prepend method above passes
3. Add a test case for testing prepend works for strings
4. A test case added for testing the fetch method, which fails
5. Correct implementation of fetch and other test cases for prepend
6. New test case for testing prepend and fetch, which passes
7. Test case for append that fails
8. Prepend followed by append passes but test append fails. Why?
9. Fix the append method
10. Refactoring to add a test case for a method called locateNode and the empty method in the class
11. Correct implementation of locateNode with appropriate test cases, which pass
12. Refactor all methods using the locateNode method.