Sort elements by frequency using STL

Given an array of integers, sort the array according to frequency of elements. If frequencies of two elements are same, print them 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

We have discussed different approaches in below posts : Sort elements by frequency | Set 1 Sort elements by frequency | Set 2 We can solve this problem using map and pairs. Initially we create a map such that map[element] = freq. Once we are done building the map, we create an array of pairs. A pair which stores elements and their corresponding frequency will be used for the purpose of sorting. We write a custom compare function which compares two pairs firstly on the basis of freq and if there is a tie on the basis of values. 

Below is its c++ implementation : 

CPP

// C++ program to sort elements by frequency using
// STL
#include <bits/stdc++.h>
using namespace std;

// function to compare two pairs for inbuilt sort
bool compare(pair<int,int> &p1,
            pair<int, int> &p2)
{
    // If frequencies are same, compare
    // values
    if (p1.second == p2.second)
        return p1.first < p2.first;
    return p1.second > p2.second;
}

// function to print elements sorted by freq
void printSorted(int arr[], int n)
{
    // Store items and their frequencies
    map<int, int> m;
    for (int i = 0; i < n; i++)
        m[arr[i]]++;

    // no of distinct values in the array
    // is equal to size of map.
    int s = m.size();

    // an array of pairs
    pair<int, int> p[s];

    // Fill (val, freq) pairs in an array
    // of pairs.
    int i = 0;
    for (auto it = m.begin(); it != m.end(); ++it)
        p[i++] = make_pair(it->first, it->second);

    // sort the array of pairs using above
    // compare function.
    sort(p, p + s, compare);

    cout << "Elements sorted by frequency are: ";
    for (int i = 0; i < s; i++)
    {
        int freq = p[i].second;
        while (freq--)
            cout << p[i].first << " ";
    }
}

// driver program
int main()
{
    int arr[] = {2, 3, 2, 4, 5, 12, 2, 3,
                3, 3, 12};
    int n = sizeof(arr)/ sizeof(arr[0]);
    printSorted(arr, n);
    return 0;
}

Java

// Java program to sort elements by frequency using
import java.util.*;

public class Main
{

    // function to compare two pairs for inbuilt sort
    static boolean compare(Map.Entry<Integer, Integer> p1,
                           Map.Entry<Integer, Integer> p2)
    {
        // If frequencies are same, compare values
        if (p1.getValue().equals(p2.getValue()))
            return p1.getKey() < p2.getKey();
        return p1.getValue() > p2.getValue();
    }

    // function to print elements sorted by freq
    static void printSorted(int[] arr, int n)
    {
        // Store items and their frequencies
        Map<Integer, Integer> m
            = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++)
            m.put(arr[i], m.getOrDefault(arr[i], 0) + 1);

        // no of distinct values in the array
        // is equal to size of map.
        int s = m.size();

        // an array of Map.Entry pairs
        List<Map.Entry<Integer, Integer> > list
            = new ArrayList<Map.Entry<Integer, Integer> >(
                m.entrySet());

        // sort the list of Map.Entry pairs using above
        // compare function.
        Collections.sort(
            list, (p1, p2) -> compare(p1, p2) ? -1 : 1);

        System.out.print(
            "Elements sorted by frequency are: ");
        for (Map.Entry<Integer, Integer> entry : list) {
            int freq = entry.getValue();
            while (freq-- > 0)
                System.out.print(entry.getKey() + " ");
        }
    }

    // driver program
    public static void main(String[] args)
    {
        int[] arr = { 2, 3, 2, 4, 5, 12, 2, 3, 3, 3, 12 };
        int n = arr.length;
        printSorted(arr, n);
    }
}

Output

Elements sorted by frequency are: 3 3 3 3 2 2 2 12 12 4 5 

Time Complexity : O(n Log n) 

Space Complexity: O(n)
The above algorithm requires O(n) space for the hash map and the array of pairs.

This article is contributed by Aditi Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.