Arayslist/list

What is a List

is an interface for ordered collections with a stable indexing order
allows duplicate elements to be inserted and provide a predictable iteration order
Has two implementations: ArrayList and LinkedList

What is an ArrayList

backed by a static array that has a fixed size.
it is a dynamic array
- The size of the array is fixed on creation, but extended on add/remove etc operations, by creating a new array of greater length and copying the elements
- The new size of the array is a geometric pattern, to allow for amortization of copying elments
Entries can be added to the array up to the maximum size of the backing array.
- When the backing array is full, the class will allocate a new, larger array and copy the old values.
- This affects performance
- An ArrayList is initially backed by an empty array. On the first addition to the ArrayList a backing array of capacity 10 is allocated.
- We can prevent this resizing behavior by passing our preferred initial capacity value to the constructor.
```
List<String> list = new ArrayList<>(1_000_000);
```
- Initializing the size on creation, improves insertion performance

Difference between Arrays and ArrayList

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Operations on ArrayList

Creation

Creating new ArrayList

List<Type> anArrayList = new ArrayList<>();

Creating an ArrayList with data, ie list of Integers
```
List<Integer> integerList = Arrays.asList(1,2,3,4);
```

General

Check if empty
- set up an empty list: List<Integer> anArrayList = new ArrayList<>();
```
anArrayList.isEmpty();
```
- will return true if anArrayList is empty.
Clear a list
- integerList.clear() removes all elements from list
Size of ArrayList
```
int sizeOfList = integerList.size();
```
Retrieving an element using its index
```
intgerList.get(<index>);
```
- integerList.get(0) returnz 1
- integerList.get(3) returnz 4
- integerList.get(sizeOfList - 1) returnz 4
- integerList.get(5) throws ArrayIndexOutOfBoundsException

Add single element to list

List<Integer> anArrayList = new ArrayList<>();

- `anArrayList.add(5)` return a boolean
- `System.out.println(anArrayList);` return [5]
  - ArrayList has toString overriden and thus shows the contents of a list instead of the object

Add without mutating

List<Integer> anArrayListWithNewElementAtTheEnd = Stream.concat(anArrayList.stream(), Stream.of(11)).collect(Collectors.toList());

Can add as many elements in Stream.of(...) if wanted to
An alternative using flatMap:

List<Integer> anArrayListWithNewElementAtTheEnd = Stream.of(anArrayList, Arrays.asList(12))
    .flatMap(Collection::stream).collect(Collectors.toList());

Can add as many elements in Arrays.asList(...) if wanted to
Can use Collections.singletonList instead of Arrays.asList(...)

Add multiple elements to list
- Add to an array of integers to the array:
  anArrayList.addAll(Arrays.asList(1, 2, 3, 4));
- System.out.println(anArrayList); returns [5, 1, 2, 3, 4] as anArrayList already has 5 in it
  - Also addAll adds to the end of the list

Replace an element at index

setup list: List<Integer> integerList = Arrays.asList(1,2,3,4);

integerList.set(1,10);

- This will mutate the integerList
- will return [1,10,3,4]

more code but does not mutate original list

List<Integer> listThatHasChangedElement = IntStream.range(0, anArrayList.size()).
      mapToObj(replaceElementAtIndex(anArrayList, 1, 4)).
      collect(Collectors.toList());

// Where replaceElementAtIndex checks the index in anArrayList to see if it matches  with indexOfElementToReplaceand replaces it with replacementElement
private static IntFunction<Integer> replaceElementAtIndex(List<Integer> anArrayList, int indexOfElementToReplace, int replacementElement) {
      return index -> index == indexOfElementToReplace ? replacementElement : anArrayList.get(index);
  }

Where 1 is the index to find element to replace
where 4 is the replacement
This will return [1,4,3,4]

Element exist in list

anArrayList.contains(<element>);

- <element> must be of same type of those in anArrayList
- `anArrayList.contains(2);` will return true as 2 is in the list
- `anArrayList.contains(10);` will return false as 10 is not in the list

anArrayList.containsAll(<arraylist of elements>);

- <arraylist of elements> must all be of same type of those in anArrayList
- `anArrayList.containsAll(Arrays.asList(1, 5));` will return true as 1 and 5 are in the list
- `anArrayList.containsAll(Arrays.asList(1, 8));` will return false as 8 is not in the list

Find index of an element

int i = anArrayList.indexOf(3);

- will return 3
- If the arraylist was  [5, 3, 2, 3, 4], then `int i = anArrayList.indexOf(3);` will return the index of the first element it matches with, which is 1

Reverse a list

Collections.reverse(anArrayList);

- This will mutate the original list, not great

List<Integer> sortedReversedOrderList = anArrayList.stream().sorted(Collections.reverseOrder()).collect(Collectors.toList());

- Can use the comparator, Collections.reverseOrder(), as list is of Integer types.

Sort list

Collections.sort(anArrayList);

- This will mutate the original list, not great
- Uses natural ordering, as Arraylist implements comparable
- If list contained elements of some user defined object, and does not have some natural ordering (does not implement comparable). then it will not sort and throw an exception with message `java.lang.ClassCastException: com.blah.Blah cannot be cast to java.lang.Comparable`. Thus needs to implement comparable
  - But can use a comparator:
  ```java
  Collections.sort(anArrayList, (x1,x2) -> x1.field - x2.field);
  ```
    - Where field is some type of some number.
    - instead of `(x1,x2) -> x1.field - x2.field` use this `Comparator.comparingInt(x -> x.id)`

List<Integer> sortedList = anArrayList.stream().sorted().collect(Collectors.toList());

Does not mutate the original list
Uses natural ordering, as Arraylist implements comparable

sublist of list

if anArrayList is [4,3,2,1,5]

List<Integer> integers = anArrayList.subList(0, 1);

- This will create a new list, with elements from anArrayList starting at 0 index up to but not including 1st index. This will return only a list of one element

List<Integer> integers = anArrayList.subList(2, 4);

- This will return [2,1]

List<Integer> integers = anArrayList.subList(0, 6);

- This will throw `java.lang.IndexOutOfBoundsException: toIndex = 6` as the list is only size 5

Remove an element

Two ways of doing this, either by remove an element using the index, or getting the first element matched.

anArrayList.remove(0)

- removes element at index 0 from the list

anArrayList.remove(new Integer(5));

- removes the object from the list

Without mutating

Removing the first specific element in the list

Integer elementToRemove = 1;
List<Integer> integers = Arrays.asList(2, 4, 1, 2, 5, 1);
int elementIndex = integers.indexOf(elementToRemove);
List<Integer> listWithfirstOneRemoved = IntStream.range(0, integers.size())
      .filter(currentIndex -> currentIndex != elementIndex)
      .mapToObj(integers::get)
      .collect(Collectors.toList());

List<Integer> result = new ArrayList<>(integers); // copy the integers list
result.remove(Integer.valueOf(elementToRemove))

Removing the element at a specific index in the list

List<Integer> result = new ArrayList<>(integers); // copy the integers list
result.remove(integers.indexOf(elementToRemove))

Learning