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

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  • Difficulty Level : Medium
  • Last Updated : 01 Sep, 2022
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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. 

mapinterface

 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)


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