Hash Sets in Java

Hash Sets are collections of unique items i.e no element can be repeated.

Hash Sets were designed to give us a way of adding and looking for unique values in a collection in a quick manner.

The values in a Hash Set can be either simple primitives like strings or integers as well as more complex object types like object literals or arrays.

In Java, the Hash Set is implemented as a HashSet object.

Creation:

To create a new empty HashSet that holds Strings, you use the following syntax:

Set<String> mySet = new HashSet<>();

Creating an empty Set takes O(1) time.

Constructing from iterable

The Set constructor also accepts an optional iterable object. If you pass an iterable object to the Set constructor, all the unique elements of the object will be added to the new set:

List<String> elements = new ArrayList<>(List.of("b", "c", "c", "d"));
Set<String> mySet = new HashSet<>(elements);

// mySet contains {"b", "c", "d"}

This takes O(n) time, where n is the number of elements in the iterable object.

Adding elements:

To add an element to the set, you use the add() method:

List<String> elements = new ArrayList<>(List.of("b", "c", "c", "d"));
Set<String> mySet = new HashSet<>(elements);

// Adding new elements
mySet.add("x");

// Adding an existing element
mySet.add("d"); // nothing happens

// mySet contains {"b", "c", "d", "x"}

The Set first checks if that element already exists and if so, it does nothing and also doesn't raise an error.

Adding an element to a Set takes O(1) time.

Checking if a value exists:

To check if a Set has a specific element, you use the contains() method which returns true if the set contains the element, otherwise, it returns false.

List<String> elements = new ArrayList<>(List.of("b", "c", "c", "d"));
Set<String> mySet = new HashSet<>(elements);

boolean exists = mySet.contains("c"); // exists is true

exists = mySet.contains("x"); // exists becomes false

Checking if a value exists in a Set takes O(1) time.

Removing elements:

To delete a specified element from a set, you use the remove() method:

List<String> elements = new ArrayList<>(List.of("b", "c", "c", "d"));
Set<String> mySet = new HashSet<>(elements);

// Deleting elements
mySet.remove("c");

// Deleting a non-existent element
mySet.remove("z"); // does nothing

// mySet contains {"b", "d"}

The Set first checks if that key exists and if not, it does nothing and also doesn't raise an error, so it's safe to delete non-existing elements.

Removing an element from a Set takes O(1) time.

Iterating over the Set values:

If we want to iterate over values of the Set, we can use the for loop along with the : operator:

List<String> elements = new ArrayList<>(List.of("b", "c", "c", "d"));
Set<String> mySet = new HashSet<>(elements);

for (String val : mySet) {
    System.out.println(val);
}

// This will print the following:
// b
// c
// d

Iterating over a Set takes O(n) time.

Space Complexity

A Set uses O(n) space, where n is the number of elements existing in the Set.

Assignment
Follow the Coding Tutorial and let's play with some Hash Sets.

Hint
Look at the examples above if you get stuck.