Sets: interview questions & scenarios

Model answer

Use a HashSet for the membership part, but normalize the value before putting it in the set. If the business rule says " Mouse ", "mouse" and "MOUSE" are the same product name, trim the text and apply a locale-stable case rule, usually toLowerCase(Locale.ROOT).

import java.util.HashSet;
import java.util.List;
import java.util.Locale;
import java.util.Set;

public class ProductNameDedup {
    public static void main(String[] args) {
        List<String> rows = List.of(" Mouse ", "mouse", "Keyboard");
        Set<String> names = new HashSet<>();

        for (String raw : rows) {
            String normalized = raw.trim().toLowerCase(Locale.ROOT);
            if (!normalized.isEmpty()) {
                names.add(normalized);
            }
        }

        System.out.println(names);
    }
}

If the UI must show products in the order they first appeared in the file, switch the concrete type to LinkedHashSet. You still get expected O(1) membership, but iteration is first-seen order. Do not rely on HashSet iteration order, even if it looks stable in a small test.

Model answer

Use a TreeSet if the collection must stay unique and sorted as you add IDs. It is backed by a tree, so add, remove and contains are O(log n), and iteration is ascending according to natural order or the comparator you pass.

import java.util.NavigableSet;
import java.util.TreeSet;

public class SortedStudentIdsExample {
    public static void main(String[] args) {
        NavigableSet<Integer> ids = new TreeSet<>();
        ids.add(1042);
        ids.add(1007);
        ids.add(1042);
        ids.add(1099);

        System.out.println(ids);              // [1007, 1042, 1099]
        System.out.println(ids.ceiling(1050)); // 1099
    }
}

If you only sort once at the end, a HashSet plus a final sort can be cheaper. But when the invariant is "always unique and ordered", TreeSet states that invariant directly.