How do you find duplicate elements in a List in Java?

Answer

Find Duplicate Elements in a List

Method 1: Using HashSet (Most Efficient — O(n))

Java

public static List<String> findDuplicates(List<String> list) {
    Set<String> seen = new HashSet<>();
    List<String> duplicates = new ArrayList<>();

    for (String item : list) {
        if (!seen.add(item)) {  // add() returns false if already exists
            if (!duplicates.contains(item)) {  // avoid adding duplicate twice
                duplicates.add(item);
            }
        }
    }
    return duplicates;
}

List<String> input = Arrays.asList("Alice", "Bob", "Alice", "Charlie", "Bob", "Dave");
System.out.println(findDuplicates(input));  // [Alice, Bob]

Method 2: Using HashMap (Get Count of Each Element)

Java

public static Map<String, Integer> getDuplicatesWithCount(List<String> list) {
    Map<String, Integer> countMap = new HashMap<>();

    // Count occurrences
    for (String item : list) {
        countMap.put(item, countMap.getOrDefault(item, 0) + 1);
    }

    // Filter entries with count > 1
    Map<String, Integer> duplicates = new LinkedHashMap<>();
    for (Map.Entry<String, Integer> entry : countMap.entrySet()) {
        if (entry.getValue() > 1) {
            duplicates.put(entry.getKey(), entry.getValue());
        }
    }
    return duplicates;
}

Map<String, Integer> result = getDuplicatesWithCount(input);
// {Alice=2, Bob=2}
result.forEach((k, v) -> System.out.println(k + " appears " + v + " times"));

Method 3: Using Java 8 Streams

Java

public static List<String> findDuplicatesWithStream(List<String> list) {
    Set<String> seen = new HashSet<>();
    return list.stream()
               .filter(item -> !seen.add(item))
               .distinct()
               .collect(Collectors.toList());
}

// Get count map with streams
Map<String, Long> frequencyMap = list.stream()
    .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

// Only duplicates (count > 1)
frequencyMap.entrySet().stream()
    .filter(e -> e.getValue() > 1)
    .forEach(e -> System.out.println(e.getKey() + ": " + e.getValue()));

Method 4: Using Collections.frequency()

Java

public static List<String> findDuplicatesFrequency(List<String> list) {
    List<String> duplicates = new ArrayList<>();
    for (String item : list) {
        if (Collections.frequency(list, item) > 1 && !duplicates.contains(item)) {
            duplicates.add(item);
        }
    }
    return duplicates;
}
// Note: O(n²) — less efficient than HashSet approach

Remove Duplicates (Bonus)

Java

// Keep only unique elements — insertion order preserved
List<String> unique = new ArrayList<>(new LinkedHashSet<>(input));
System.out.println(unique); // [Alice, Bob, Charlie, Dave]

With Integers

Java

List<Integer> numbers = Arrays.asList(1, 2, 3, 2, 4, 3, 5, 1);

// Find duplicates
Set<Integer> seen = new HashSet<>();
Set<Integer> duplicates = new LinkedHashSet<>();
for (int n : numbers) {
    if (!seen.add(n)) duplicates.add(n);
}
System.out.println(duplicates); // [2, 3, 1]

In Automation Testing

Java

// Find duplicate options in a dropdown
List<String> options = driver.findElements(By.tagName("option"))
                             .stream()
                             .map(WebElement::getText)
                             .collect(Collectors.toList());

Set<String> seen = new HashSet<>();
List<String> duplicates = options.stream()
    .filter(o -> !seen.add(o))
    .collect(Collectors.toList());

assertTrue(duplicates.isEmpty(), "Dropdown has duplicate options: " + duplicates);