Java List, Keep First Item, Remove Second Duplicate [duplicate]

Question

I have a list from a legacy system. Its in a List. Sometimes it may have duplicates.

How do I remove second duplicate (or multiple) while keeping first one? For example, keep 3-furniture and remove 3-TV. Removal is based on alphabetization.

ProductId	ProductTitle	ProductDescription
1	car	used for driving
2	book	reading books
3	furniture	home decorations
3	TV	TV entertainment
4	jewelry	fashion item

@Data
public class Product {
    private long productId;
    private String productTitle;
    private String productDescription;
 }

What do you mean "based on alphabetization"? Do you want to keep the first one sharing the same ProductId based on their ProductTitle alphabetical ascending sort? — Nikolas
– Nikolas, Commented Mar 8, 2023 at 3:50
based on productTitle alphabetization, keep 3-f, remove 3t, cc @NikolasCharalambidis — mattsmith5
– mattsmith5, Commented Mar 8, 2023 at 3:51

Nikolas · Accepted Answer · 2023-03-08 04:05:49Z

You can achieve it in two steps.

Group to a Map where the productId is a key and the product with the lowest alphabetical productTitle using Collectors.maxBy as a downstream collector.

Map<Long, Optional<Product>> productMap = products.stream().collect(
    Collectors.groupingBy(
        Product::getProductId,
        Collectors.maxBy(Comparator.comparing(Product::getProductTitle))));

productMap.forEach((id, opt) -> System.out.println(id + " = " + opt));

1 = Optional[Product(productId=1, productTitle=car, productDescription=used for driving)]
2 = Optional[Product(productId=2, productTitle=book, productDescription=reading books)]
3 = Optional[Product(productId=3, productTitle=furniture, productDescription=home decorations)]
4 = Optional[Product(productId=4, productTitle=jewelry, productDescription=fashion item)]

Extract each Product from Optional in the Map values to a new List<Product>.

List<Product> filteredProducts = productMap.values().stream()
    .filter(Optional::isPresent)
    .map(Optional::get)
    .toList();

filteredProducts.forEach(System.out::println);

Product(productId=1, productTitle=car, productDescription=used for driving)
Product(productId=2, productTitle=book, productDescription=reading books)
Product(productId=3, productTitle=furniture, productDescription=home decorations)
Product(productId=4, productTitle=jewelry, productDescription=fashion item)

sendon1982 · Accepted Answer · 2023-03-08 04:10:29Z

Hi you can easily do it in Java with PriorityQueue and time complexity is O(nlogn)

    public static List<Product> filterProduct(List<Product> products) {
        PriorityQueue<Product> pq = new PriorityQueue<>((a, b) -> {
            if (a.getProductId() == b.getProductId()) {
                return a.getProductTitle().toLowerCase().compareTo(b.getProductTitle().toLowerCase());
            }

            return Long.compare(a.getProductId(), b.getProductId());
        });

        pq.addAll(products);

        Set<Long> existingIds = new HashSet<>();

        List<Product> res = new ArrayList<>();

        while (!pq.isEmpty()) {
            Product product = pq.poll();
            if (!existingIds.contains(product.getProductId())) {
                res.add(product);
                existingIds.add(product.getProductId());
            }
        }

        return res;
    }

Collectives™ on Stack Overflow

Java List, Keep First Item, Remove Second Duplicate [duplicate]

2 Answers 2

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Linked

Related