Sort elements by frequency | Set 5 (using Java Map)

Given an integer array, sort the array according to the frequency of elements in decreasing order, if the frequency of two elements are same then sort in increasing order 

Examples:

Input: arr[] = {2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12}
Output: 3 3 3 3 2 2 2 12 12 4 5
Explanation :
No. Freq
2  : 3
3  : 4
4  : 1
5  : 1
12 : 2

Input: arr[] = {4, 4, 2, 2, 2, 2, 3, 3, 1, 1, 6, 7, 5}
Output: 2 2 2 2 1 1 3 3 4 4 5 6 7

Different approaches have been discussed in below posts: 

Sort elements by frequency | Set 1
Sort elements by frequency | Set 2
Sorting Array Elements By Frequency | Set 3 (Using STL)
Sort elements by frequency | Set 4 (Efficient approach using hash)

Approach:

Java Map has been used in this set to solve the problem. 

The java.util.Map interface represents a mapping between a key and a value. The Map interface is not a subtype of the Collection interface. Therefore it behaves a bit different from the rest of the collection types. 

In the below program:

• Get the element with its count in a Map
• By using the Comparator Interface, compare the frequency of an elements in a given list.
• Use this comparator to sort the list by implementing Collections.sort() method.
• Print the sorted list.

Implementation:

Java

import java.util.*;

public class GFG {

    // Driver Code
    public static void main(String[] args)
    {

        // Declare and Initialize an array
        int[] array = { 4, 4, 2, 2, 2, 2, 3, 3, 1, 1, 6, 7, 5 };

        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> outputArray = new ArrayList<>();

        // Assign elements and their count in the list and map
        for (int current : array) {
            int count = map.getOrDefault(current, 0);
            map.put(current, count + 1);
            outputArray.add(current);
        }

        // Compare the map by value
        SortComparator comp = new SortComparator(map);

        // Sort the map using Collections CLass
        Collections.sort(outputArray, comp);

        // Final Output
        for (Integer i : outputArray) {
            System.out.print(i + " ");
        }
    }
}

// Implement Comparator Interface to sort the values
class SortComparator implements Comparator<Integer> {
    private final Map<Integer, Integer> freqMap;

    // Assign the specified map
    SortComparator(Map<Integer, Integer> tFreqMap)
    {
        this.freqMap = tFreqMap;
    }

    // Compare the values
    @Override
    public int compare(Integer k1, Integer k2)
    {

        // Compare value by frequency
        int freqCompare = freqMap.get(k2).compareTo(freqMap.get(k1));

        // Compare value if frequency is equal
        int valueCompare = k1.compareTo(k2);

        // If frequency is equal, then just compare by value, otherwise -
        // compare by the frequency.
        if (freqCompare == 0)
            return valueCompare;
        else
            return freqCompare;
    }
}

Output

2 2 2 2 1 1 3 3 4 4 5 6 7 

Time Complexity: O(n Log n)

Space complexity: The space complexity of the above code is O(n) as we are using a hashmap and an arraylist of size n.