How do you sort an array based on count of occurrences in ascending order

Problem statement:
Given an array of integer, you have to sort based on count of occurrences in ascending order?

1. import java.util.*;
2. public class SortArrayInDesc {
3.     public static void main(String[] args) {
4.         int a[] = {2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 1, 1, 1};
5.         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
6.         for (int i = 0; i < a.length; i++) {
7.             if (map.containsKey(a[i])) {
8.               map.put(a[i], map.get(a[i]) + 1);
9.             } else {
10.                 map.put(a[i], 1);
11.             }
12.         }
13.         ValueComparator<Integer, Integer> comparator = new ValueComparator<Integer, Integer>(map);
14.         TreeMap<Integer, Integer> sortedMap = new TreeMap<Integer, Integer>(comparator);
15.         sortedMap.putAll(map);
16.         ArrayList<Integer> lists = new ArrayList<Integer>();
17.         for (Integer i : sortedMap.keySet()) {
18.             for (int c = 0; c < sortedMap.get(i); c++) {
19.                 lists.add(i);
20.             }
21.         }
22.         System.out.println(lists.toString());
23.     }
24. }

1. import java.util.Comparator;
2. import java.util.Map;
3. public class ValueComparator<T1, T2 extends Comparable<T2>> implements Comparator<T1> {
4.     Map<T1, T2> map;
5.     public ValueComparator(Map<T1, T2> map) {
6.         this.map = map;
7.     }
8.     @Override
9.     public int compare(T1 k1, T1 k2) {
10.         T2 val1 = map.get(k1);
11.         T2 val2 = map.get(k2);
12. 
    
13.         return val1.compareTo(val2);
14.     }
15. }

Output:

[5, 1, 1, 1, 2, 2, 2, 2, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4]