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;
 }

Answer 1

You can achieve it in two steps.

1. 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)]
   

2. 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)
   

Answer 2

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;
    }