Last updated: September 25, 2023
In this short tutorial, we’ll see how to find the n most frequent elements in a Java array.
We can use a HashMap to count the occurrences of each element and a PriorityQueue to prioritize elements based on their count.
This allows us to find the n most frequent elements in the array efficiently:
public static List<Integer> findByHashMapAndPriorityQueue(Integer[] array, int n) {
Map<Integer, Integer> countMap = new HashMap<>();
// For each element i in the array, add it to the countMap and increment its count
for (Integer i : array) {
countMap.put(i, countMap.getOrDefault(i, 0) + 1);
}
// Create a max heap (priority queue) that will prioritize elements with higher counts.
PriorityQueue<Integer> heap = new PriorityQueue<>((a, b) -> countMap.get(b) - countMap.get(a));
// Add all the unique elements in the array to the heap.
heap.addAll(countMap.keySet());
List<Integer> result = new ArrayList<>();
for (int i = 0; i < n && !heap.isEmpty(); i++) {
// Poll the highest-count element from the heap and add it to the result list.
result.add(heap.poll());
}
return result;
}Let’s test our methods:
Integer[] inputArray = {1, 2, 3, 2, 2, 1, 4, 5, 6, 1, 2, 3};
Integer[] outputArray = {2, 1, 3};
assertThat(findByHashMapAndPriorityQueue(inputArray, 3)).containsExactly(outputArray);We can also use the Stream API to create a Map to count the occurrences of each element, sort the entries in the map by frequency in descending order, and then extract the n most frequent elements:
public static List<Integer> findByStream(Integer[] arr, int n) {
return Arrays.stream(arr).collect(Collectors.groupingBy(i -> i, Collectors.counting()))
.entrySet().stream()
.sorted(Collections.reverseOrder(Map.Entry.comparingByValue()))
.map(Map.Entry::getKey)
.limit(n)
.collect(Collectors.toList());
}Alternatively, we can use a TreeMap and create a custom Comparator that compares the values in the Map.Entry objects in the map by frequency in descending order:
public static List<Integer> findByTreeMap(Integer[] arr, int n) {
// Create a TreeMap and use a reverse order comparator to sort the entries by frequency in descending order
Map<Integer, Integer> countMap = new TreeMap<>(Collections.reverseOrder());
for (int i : arr) {
countMap.put(i, countMap.getOrDefault(i, 0) + 1);
}
// Create a list of the map entries and sort them by value (i.e. by frequency) in descending order
List<Map.Entry<Integer, Integer>> sortedEntries = new ArrayList<>(countMap.entrySet());
sortedEntries.sort((e1, e2) -> e2.getValue().compareTo(e1.getValue()));
// Extract the n most frequent elements from the sorted list of entries
List<Integer> result = new ArrayList<>();
for (int i = 0; i < n && i < sortedEntries.size(); i++) {
result.add(sortedEntries.get(i).getKey());
}
return result;
}In summary, we’ve learned different ways to find the n most frequent elements in a Java array.
The example code from this article can be found over on GitHub.
