Sort elements by frequency | Set 4 (Efficient approach using hash)

Print the elements of an array in the decreasing frequency if 2 numbers have the same frequency then print the one which came first.

Examples:

Input : arr[] = {2, 5, 2, 8, 5, 6, 8, 8}
Output : arr[] = {8, 8, 8, 2, 2, 5, 5, 6}

Input : arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8}
Output : arr[] = {8, 8, 8, 2, 2, 5, 5, 6, -1, 9999999}
 

We have discussed different approaches in below posts : 
Sort elements by frequency | Set 1
Sort elements by frequency | Set 2
Sorting Array Elements By Frequency | Set 3 (Using STL)
All of the above approaches work in O(n Log n) time where n is total number of elements. In this post, a new approach is discussed that works in O(n + m Log m) time where n is total number of elements and m is total number of distinct elements.

The idea is to use hashing. 

1. We insert all elements and their counts into a hash. This step takes O(n) time where n is number of elements.
2. We copy the contents of hash to an array (or vector) and sort them by counts. This step takes O(m Log m) time where m is total number of distinct elements.
3. For maintaining the order of elements if the frequency is the same, we use another hash which has the key as elements of the array and value as the index. If the frequency is the same for two elements then sort elements according to the index.

The below image is a dry run of the above approach:

We do not need to declare another map m2, as it does not provide the proper expected result for the problem.

instead, we need to just check for the first values of the pairs sent as parameters in the sortByVal function.

Below is the implementation of the above approach:

C++

//     CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Used for sorting by frequency. And if frequency is same,
// then by appearance
bool sortByVal(const pair<int, int>& a,
                      const pair<int, int>& b)
{

   // If frequency is same then sort by index
   if (a.second == b.second)
       return a.first < b.first;

   return a.second > b.second;
}

// function to sort elements by frequency
vector<int>sortByFreq(int a[], int n)
{

   vector<int>res;

   unordered_map<int, int> m;

   vector<pair<int, int> > v;

   for (int i = 0; i < n; ++i) {

       // Map m is used to keep track of count
       // of elements in array
       m[a[i]]++;
   }

   // Copy map to vector
   copy(m.begin(), m.end(), back_inserter(v));

   // Sort the element of array by frequency
   sort(v.begin(), v.end(), sortByVal);

   for (int i = 0; i < v.size(); ++i)
      while(v[i].second--)
      {
              res.push_back(v[i].first);
      }

   return res;
}

// Driver program
int main()
{

   int a[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
   int n = sizeof(a) / sizeof(a[0]);
   vector<int>res;
   res = sortByFreq(a, n);

   for(int i = 0;i < res.size(); i++)
         cout<<res[i]<<" ";

   return 0;

}

Python3

# Used for sorting by frequency. And if frequency is same,
# then by appearance
from functools import cmp_to_key

def sortByVal(a,b):

    # If frequency is same then sort by index
    if (a[1] == b[1]):
        return a[0] - b[0]

    return b[1] - a[1]

# function to sort elements by frequency
def sortByFreq(a, n):
    res = []
    m = {}
    v = []

    for i in range(n):

        # Map m is used to keep track of count
        # of elements in array
        if(a[i] in m):
            m[a[i]] = m[a[i]]+1
        else:
            m[a[i]] = 1

    for key,value in m.items():
        v.append([key,value])

    # Sort the element of array by frequency
    v.sort(key = cmp_to_key(sortByVal))

    for i in range(len(v)):
        while(v[i][1]):
            res.append(v[i][0])
            v[i][1] -= 1

    return res


# Driver program

a = [ 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 ]
n = len(a)
res = []
res = sortByFreq(a, n)

for i in range(len(res)):
    print(res[i],end = " ")


# This code is contributed by shinjanpatra

Java

// Java program for the above approach
import java.util.*;

// Used for sorting by frequency. And if frequency is same,
// then by appearance
class SortByValue implements Comparator<Map.Entry<Integer, Integer> >
{
    // Used for sorting in descending order of values
    public int compare(Map.Entry<Integer, Integer> o1,
                       Map.Entry<Integer, Integer> o2)
    {
        // If frequency is same then sort by index
        if (o1.getValue() == o2.getValue())
            return o1.getKey() - o2.getKey();
        return o2.getValue() - o1.getValue();
    }
}

class GFG
{
    // Function to sort elements by frequency
    static Vector<Integer> sortByFreq(int a[], int n)
    {
        // Map to store the frequency of the elements
        HashMap<Integer, Integer> m = new HashMap<>();

        // Vector to store the sorted elements
        Vector<Integer> v = new Vector<>();

        // Insert elements and their frequency in the map
        for (int i = 0; i < n; i++)
        {
            int x = a[i];
            if (m.containsKey(x))
                m.put(x, m.get(x) + 1);
            else
                m.put(x, 1);
        }

        // Copy map to vector
        Vector<Map.Entry<Integer, Integer> > v1 =
            new Vector<>(m.entrySet());

        // Sort the vector elements by frequency
        Collections.sort(v1, new SortByValue());

        // Traverse the vector and insert elements
        // in the vector v
        for (int i = 0; i < v1.size(); i++)
            for (int j = 0; j < v1.get(i).getValue(); j++)
                v.add(v1.get(i).getKey());

        return v;
    }

    // Driver program
    public static void main(String[] args)
    {
        int a[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
        int n = a.length;
        Vector<Integer> v = sortByFreq(a, n);

        // Print the elements of vector
        for (int i = 0; i < v.size(); i++)
            System.out.print(v.get(i) + " ");
    }
}

Javascript

<script>

//     JavaScript program for the above approach


// Used for sorting by frequency. And if frequency is same,
// then by appearance
function sortByVal(a,b)
{

// If frequency is same then sort by index
if (a[1] == b[1])
    return a[0] - b[0];

return b[1] - a[1];
}

// function to sort elements by frequency
function sortByFreq(a, n)
{

let res = [];

let m = new Map();

let v = [];

for (let i = 0; i < n; ++i) {

    // Map m is used to keep track of count
    // of elements in array
    if(m.has(a[i]))
        m.set(a[i],m.get(a[i])+1);
    else
        m.set(a[i],1);
}

for(let [key,value] of m){
    v.push([key,value]);
}

// Sort the element of array by frequency
v.sort(sortByVal)

for (let i = 0; i < v.length; ++i)
    while(v[i][1]--)
    {
        res.push(v[i][0]);
    }

    return res;
}

// Driver program

let a = [ 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 ];
let n = a.length;
let res = [];
res = sortByFreq(a, n);

for(let i = 0;i < res.length; i++)
    document.write(res[i]," ");


// This code is contributed by shinjanpatra

</script>

C#

using System;
using System.Collections.Generic;
using System.Linq;

// Used for sorting by frequency. And if frequency is same,
// then by appearance
class SortByValue : IComparer<KeyValuePair<int, int>>
{
    // Used for sorting in descending order of values
    public int Compare(KeyValuePair<int, int> o1, KeyValuePair<int, int> o2)
    {
        // If frequency is same then sort by index
        if (o1.Value == o2.Value)
            return o1.Key - o2.Key;
        return o2.Value - o1.Value;
    }
}

class GFG
{
    // Function to sort elements by frequency
    static List<int> sortByFreq(int[] a, int n)
    {
        // Dictionary to store the frequency of the elements
        Dictionary<int, int> m = new Dictionary<int, int>();

        // List to store the sorted elements
        List<int> res = new List<int>();

        // Insert elements and their frequency in the dictionary
        for (int i = 0; i < n; i++)
        {
            int x = a[i];
            if (m.ContainsKey(x))
                m[x]++;
            else
                m.Add(x, 1);
        }

        // Copy dictionary to list
        List<KeyValuePair<int, int>> v =
            new List<KeyValuePair<int, int>>(m);

        // Sort the list elements by frequency
        v.Sort(new SortByValue());

        // Traverse the list and insert elements
        // in the list v
        foreach (KeyValuePair<int, int> kvp in v)
        {
            for (int i = 0; i < kvp.Value; i++)
                res.Add(kvp.Key);
        }

        return res;
    }

    // Driver program
    public static void Main(string[] args)
    {
        int[] a = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
        int n = a.Length;
        List<int> res = sortByFreq(a, n);

        // Print the elements of list
        foreach (int i in res)
            Console.Write(i + " ");
    }
}

Output

8 8 8 2 2 5 5 -1 6 9999999 

Time Complexity: O(n) + O(m Log m) where n is total number of elements and m is total number of distinct elements
Auxiliary Space: O(n)

This article is contributed by Aarti_Rathi and Ankur Singh and improved by Ankur Goel. 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. 

Simple way to sort by frequency.

The Approach:

  Here In This approach we first we store the element by there frequency in vector_pair format(Using Mapping stl map) then sort it according to frequency then reverse it and apply bubble sort to make the condition true decreasing frequency if 2 numbers have the same frequency then print the one which came first. then print the vector.

C++

#include <bits/stdc++.h>
#include<iostream>
using namespace std;

//map all the number and sort by frequency.
void the_helper(int a[],vector<pair<int,int>>&res,int n){
   map<int,int>mp;
   for(int i=0;i<n;i++)mp[a[i]]++;
   for(auto it:mp)res.push_back({it.second,it.first});
   sort(res.begin(),res.end());
}

int main() {
  int a[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8};
  vector<pair<int,int>>res;
  the_helper(a,res,10);
  reverse(res.begin(),res.end());
  for(int i=0;i<res.size();i++){
    if(res[i].first==res[i+1].first){
      for(int j=i;j<res.size();j++){
        if(res[i].second>res[j].second&&res[i].first==res[j].first){
          swap(res[i],res[j]);
        }
      }
    }
  }
  for(int i=0;i<res.size();i++){
    for(int j=0;j<res[i].first;j++)cout<<res[i].second<<" ";
  //  cout<<endl;
  }
  return 0;
}

Java

// Java program for the above approach
import java.util.*;

// pair class
class Pair{
    int first;
    int second;
    public Pair(int first, int second){
        this.first = first;
        this.second = second;
    }
}

public class Main{

    // map all the number and sort by frequency
    public static void the_helper(int[] a, List<Pair> res, int n){
        // create a new map to store the frequency of each number in the array
        Map<Integer, Integer> mp = new HashMap<>();
        for(int i = 0; i < n; i++){
            // check if the map already has the number,
              // if yes, increase its count by 1, else add it to the map with count 1
            if(mp.containsKey(a[i]))
                mp.put(a[i], mp.get(a[i])+1);
            else
                mp.put(a[i], 1);
        }
        // loop through the map entries and add them to the result list as pairs
        for(Map.Entry<Integer, Integer> entry : mp.entrySet()){
            res.add(new Pair(entry.getValue(), entry.getKey()));
        }
        // sort the result list in ascending order of
          // frequency using a lambda expression
        res.sort((x, y) -> x.first - y.first);
    }

    // driver program
    public static void main(String[] args) {
        int[] a = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8};
        List<Pair> res = new ArrayList<>();
        the_helper(a, res, 10);
        // reverse the result list to get it in descending order of frequency
        Collections.reverse(res);
        // loop through the result list and swap pairs with
          // equal frequencies if the second element of the earlier
          // pair is greater than the second element of the later pair
        for(int i = 0; i < res.size()-1; i++){
            if(res.get(i).first == res.get(i+1).first){
                for(int j = i; j < res.size(); j++){
                    if(res.get(i).second > res.get(j).second && res.get(i).first == res.get(j).first){
                        Pair temp = res.get(j);
                        res.set(j, res.get(i));
                        res.set(i, temp);
                    }
                }
            }
        }
        System.out.println();
        // loop through the result list and print each pair's
          //second element the number of times indicated by its first element
        for(Pair p : res){
            for(int i = 0; i < p.first; i++){
                System.out.print(p.second + " ");
            }
        }
    }
}

// this code is contributed by bhardwajji

Python3

# python3 program for the above approach
import collections

# map all the number and sort by frequency


def the_helper(a, res, n):
    mp = collections.defaultdict(int)
    for i in range(n):
        mp[a[i]] += 1
    for key, val in mp.items():
        res.append((val, key))
    res.sort()


# main function
if __name__ == '__main__':
    a = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8]
    res = []
    the_helper(a, res, len(a))
    res.reverse()
    for i in range(len(res) - 1):
        if res[i][0] == res[i+1][0]:
            for j in range(i+1, len(res)):
                if res[i][0] == res[j][0] and res[i][1] > res[j][1]:
                    res[i], res[j] = res[j], res[i]
    for i in range(len(res)):
        for j in range(res[i][0]):
            print(res[i][1], end=' ')
        # print()  # uncomment to print each frequency on a new line

C#

using System;
using System.Collections.Generic;
using System.Linq;

public class Program {
    //map all the number and sort by frequency.
    public static void the_helper(int[] a, List<Tuple<int,int>> res, int n) {
        Dictionary<int,int> mp = new Dictionary<int,int>();
        for (int i = 0; i < n; i++) {
            if (mp.ContainsKey(a[i])) {
                mp[a[i]]++;
            } else {
                mp[a[i]] = 1;
            }
        }
        foreach (var it in mp) {
            res.Add(new Tuple<int,int>(it.Value, it.Key));
        }
        res.Sort();
    }

    public static void Main() {
        int[] a = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
        List<Tuple<int,int>> res = new List<Tuple<int,int>>();
        the_helper(a, res, 10);
        res.Reverse();
        for (int i = 0; i < res.Count; i++) {
            if (i < res.Count - 1 && res[i].Item1 == res[i+1].Item1) {
                for (int j = i; j < res.Count; j++) {
                    if (res[i].Item2 > res[j].Item2 && res[i].Item1 == res[j].Item1) {
                        var temp = res[i];
                        res[i] = res[j];
                        res[j] = temp;
                    }
                }
            }
        }
        for(int i=0;i<res.Count;i++){
            for (int j = 0; j < res[i].Item1; j++) {
                Console.Write(res[i].Item2 + " ");
            }
        }
    }
}

Javascript

// JavaScript program for the above approach
// pair class
class pair{
    constructor(first, second){
        this.first = first;
        this.second = second;
    }
}

// map all the number and sort by frequency
function the_helper(a, res, n){
    mp = new Map();
    for(let i = 0; i<n; i++){
        if(mp.has(a[i]))
            mp.set(a[i], mp.get(a[i])+1);
        else
            mp.set(a[i], 1);
    }
    mp.forEach(function(value, key){
        res.push(new pair(value, key));
    })
    res.sort(function(a, b){
        return a.first - b.first;
    });
}

// driver program
let a = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8];
let res = [];
the_helper(a, res, 10);
res.reverse();
for(let i = 0; i < res.length-1; i++){
    if(res[i].first == res[i+1].first){
        for(let j = i; j < res.length; j++){
            if(res[i].second > res[j].second && res[i].first == res[j].first){
                let temp = res[j];
                res[j] = res[i];
                res[i] = temp;
            }
        }
    }
}
console.log("\n");
for(let i = 0; i < res.length; i++){
    for(let j = 0; j < res[i].first; j++){
        console.log(res[i].second + " ");
    }
}

// this code is contributed by Yash Agarwal(yashagarwal2852002)

Output

8 8 8 2 2 5 5 -1 6 9999999 

Time Complexity: O(n^2) I.e it take O(n) for getting the frequency sorted vector but for sorting in decreasing frequency if 2 numbers have the same frequency then print the one which came first we use bubble sort so it take O(n^2).
Auxiliary Space: O(n),for vector.