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Sorting element of an array by frequency in decreasing order

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  • Difficulty Level : Easy
  • Last Updated : 21 Apr, 2022
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Given an array arr[] of N integers. The task is to sort the array arr[] according to the frequency of elements in decreasing order. 
Note: If the frequencies of the two elements are the same, then the smaller element should come first.

Examples:  

Input: arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 } 
Output: 4 4 4 2 2 5 5 6 8

Input: arr[] = { 9, 9, 5, 8, 5 } 
Output: 5 5 9 9 8 
  

Approach: 

  • Store the frequency of all elements in array arr[].
  • For arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 }
    The frequency array of the above array is:
    freq[] = { 0, 0, 2, 0, 3, 2, 0, 1, 0, 1}
  • Traverse the frequency array and for all the element having frequency greater than 1, update the value in the array arr[] as:

freq[2] = 2 
arr[0] = 100000*freq[2] + (100000 – 2) = 299998 
freq[4] = 3 
arr[1] = 100000*freq[2] + (100000 – 4) = 399996 
freq[5] = 2 
arr[2] = 100000*freq[2] + (100000 – 5) = 299995 
freq[6] = 2 
arr[3] = 100000*freq[2] + (100000 – 6) = 199994 
freq[8] = 2 
arr[4] = 100000*freq[2] + (100000 – 2) = 199994 
Now array becomes: 
arr[] = {299998, 399996, 299995, 199994, 199992, 2, 2, 8, 5} 
 

  • Sort the array arr[] in decreasing order.
  • Traverse the array arr[] and to get the element and frequency of that element as per the updation of array element in step 2, do the following:

To get the frequency of current element: 
frequency = arr[i]/100000; 
To get the value: 
value = 100000 – arr[i]%100000 
 

For Example:  

if arr[i] = 399996 
frequency = arr[i]/100000 = 399996/100000 = 3 
value = 100000 – arr[i]%100000 = 100000 – 99996 = 4 
The element 4 is having frequency 3
 

  • For each element in arr[] find the value and frequency(say f) at each index and print the value f number of times.

Below is the implementation of the above approach:  

C++




// C++ program to sort an array in
// decreasing order of their frequency
#include "bits/stdc++.h"
using namespace std;
 
// Function that return the index
// upto all the array elements are
// updated.
int sortByFreq(int* arr, int n)
{
 
    // Initialise maxE = -1
    int maxE = -1;
 
    // Find the maximum element of
    // arr[]
    for (int i = 0; i < n; i++) {
        maxE = max(maxE, arr[i]);
    }
 
    // Create frequency array freq[]
    int freq[maxE + 1] = { 0 };
 
    // Update the frequency array as
    // per the occurrence of element in
    // arr[]
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
 
    // Initialise cnt to 0
    int cnt = 0;
 
    // Traversing freq[]
    for (int i = 0; i <= maxE; i++) {
 
        // If freq of an element is
        // greater than 0 update the
        // value of arr[] at index cnt
        // & increment cnt
        if (freq[i] > 0) {
 
            int value = 100000 - i;
            arr[cnt] = 100000 * freq[i] + value;
            cnt++;
        }
    }
 
    // Return cnt
    return cnt;
}
 
// Function that print array arr[]
// elements in sorted order
void printSortedArray(int* arr, int cnt)
{
 
    // Traversing arr[] till index cnt
    for (int i = 0; i < cnt; i++) {
 
        // Find frequency of elements
        int frequency = arr[i] / 100000;
 
        // Find value at index i
        int value = 100000 - (arr[i] % 100000);
 
        // Traversing till frequency
        // to print value at index i
        for (int j = 0; j < frequency; j++) {
            cout << value << ' ';
        }
    }
}
 
// Driver code
int main()
{
 
    int arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
 
    // Size of array arr[]
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to get cnt
    int cnt = sortByFreq(arr, n);
 
    // Sort the arr[] in decreasing order
    sort(arr, arr + cnt, greater<int>());
 
    // Function that prints elements
    // in decreasing order
    printSortedArray(arr, cnt);
 
    return 0;
}

Java




// Java program to sort an array in
// decreasing order of their frequency
import java.util.*;
 
class GFG{
  
// Function that return the index
// upto all the array elements are
// updated.
static int sortByFreq(Integer []arr, int n)
{
  
    // Initialise maxE = -1
    int maxE = -1;
  
    // Find the maximum element of
    // arr[]
    for (int i = 0; i < n; i++) {
        maxE = Math.max(maxE, arr[i]);
    }
  
    // Create frequency array freq[]
    int freq[] = new int[maxE + 1];
  
    // Update the frequency array as
    // per the occurrence of element in
    // arr[]
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
  
    // Initialise cnt to 0
    int cnt = 0;
  
    // Traversing freq[]
    for (int i = 0; i <= maxE; i++) {
  
        // If freq of an element is
        // greater than 0 update the
        // value of arr[] at index cnt
        // & increment cnt
        if (freq[i] > 0) {
  
            int value = 100000 - i;
            arr[cnt] = 100000 * freq[i] + value;
            cnt++;
        }
    }
  
    // Return cnt
    return cnt;
}
  
// Function that print array arr[]
// elements in sorted order
static void printSortedArray(Integer []arr, int cnt)
{
  
    // Traversing arr[] till index cnt
    for (int i = 0; i < cnt; i++) {
  
        // Find frequency of elements
        int frequency = arr[i] / 100000;
  
        // Find value at index i
        int value = 100000 - (arr[i] % 100000);
  
        // Traversing till frequency
        // to print value at index i
        for (int j = 0; j < frequency; j++) {
            System.out.print(value + " ");
        }
    }
}
  
// Driver code
public static void main(String[] args)
{
  
    Integer arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
  
    // Size of array arr[]
    int n = arr.length;
  
    // Function call to get cnt
    int cnt = sortByFreq(arr, n);
  
    // Sort the arr[] in decreasing order
    Arrays.sort(arr, Collections.reverseOrder());
  
    // Function that prints elements
    // in decreasing order
    printSortedArray(arr, cnt);
  
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python program to sort an array in
# decreasing order of their frequency
 
# Function that return the index
# upto all the array elements are
# updated.
def sortByFreq(arr, n):
 
    # Initialise maxE = -1
    maxE = -1;
 
    # Find the maximum element of
    # arr[]
    for i in range(n):
        maxE = max(maxE, arr[i])
 
    # Create frequency array freq[]
    freq = [0]*(maxE + 1);
 
    # Update the frequency array as
    # per the occurrence of element in
    # arr[]
    for i in range(n):
        freq[arr[i]] += 1;
     
    # Initialise cnt to 0
    cnt = 0;
 
    # Traversing freq[]
    for i in range(maxE+1):
 
        # If freq of an element is
        # greater than 0 update the
        # value of arr[] at index cnt
        # & increment cnt
        if (freq[i] > 0):
 
            value = 100000 - i;
            arr[cnt] = 100000 * freq[i] + value;
            cnt += 1;
         
    # Return cnt
    return cnt;
 
# Function that print array arr[]
# elements in sorted order
def printSortedArray(arr, cnt):
 
    # Traversing arr[] till index cnt
    for i in range(cnt):
 
        # Find frequency of elements
        frequency = arr[i] / 100000;
 
        # Find value at index i
        value = 100000 - (arr[i] % 100000);
 
        # Traversing till frequency
        # to print value at index i
        for j in range(int(frequency)):
            print(value, end=" ")
             
# Driver code
if __name__=='__main__':
 
    arr = [ 4, 4, 5, 6, 4, 2, 2, 8, 5 ]
 
    # Size of array arr[]
    n = len(arr)
 
    # Function call to get cnt
    cnt = sortByFreq(arr, n);
 
    # Sort the arr[] in decreasing order
    arr.sort(reverse = True)
 
    # Function that prints elements
    # in decreasing order
    printSortedArray(arr, cnt);
 
# This code is contributed by Princi Singh

C#




// C# program to sort an array in
// decreasing order of their frequency
using System;
 
class GFG {
 
  // Function that return the index
  // upto all the array elements are
  // updated.
  static int sortByFreq(int[] arr, int n)
  {
 
    // Initialise maxE = -1
    int maxE = -1;
 
    // Find the maximum element of
    // arr[]
    for (int i = 0; i < n; i++)
    {
      maxE = Math.Max(maxE, arr[i]);
    }
 
    // Create frequency array freq[]
    int[] freq = new int[maxE + 1];
 
    // Update the frequency array as
    // per the occurrence of element in
    // arr[]
    for (int i = 0; i < n; i++)
    {
      freq[arr[i]]++;
    }
 
    // Initialise cnt to 0
    int cnt = 0;
 
    // Traversing freq[]
    for (int i = 0; i <= maxE; i++)
    {
 
      // If freq of an element is
      // greater than 0 update the
      // value of arr[] at index cnt
      // & increment cnt
      if (freq[i] > 0)
      {
        int value = 100000 - i;
        arr[cnt] = 100000 * freq[i] + value;
        cnt++;
      }
    }
 
    // Return cnt
    return cnt;
  }
 
  // Function that print array arr[]
  // elements in sorted order
  static void printSortedArray(int[] arr, int cnt)
  {
 
    // Traversing arr[] till index cnt
    for (int i = 0; i < cnt; i++)
    {
 
      // Find frequency of elements
      int frequency = arr[i] / 100000;
 
      // Find value at index i
      int value = 100000 - (arr[i] % 100000);
 
      // Traversing till frequency
      // to print value at index i
      for (int j = 0; j < frequency; j++)
      {
        Console.Write(value + " ");
      }
    }
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
 
    // Size of array arr[]
    int n = arr.Length;
 
    // Function call to get cnt
    int cnt = sortByFreq(arr, n);
 
    // Sort the arr[] in decreasing order
    Array.Sort(arr);
    Array.Reverse(arr);
 
    // Function that prints elements
    // in decreasing order
    printSortedArray(arr, cnt);
  }
}
 
// This code is contributed by subhamahato348

Javascript




<script>
 
      // JavaScript program to sort an array in
      // decreasing order of their frequency
       
      // Function that return the index
      // upto all the array elements are
      // updated.
      function sortByFreq(arr, n) {
        // Initialise maxE = -1
        var maxE = -1;
 
        // Find the maximum element of
        // arr[]
        for (var i = 0; i < n; i++) {
          maxE = Math.max(maxE, arr[i]);
        }
 
        // Create frequency array freq[]
        var freq = new Array(maxE + 1).fill(0);
 
        // Update the frequency array as
        // per the occurrence of element in
        // arr[]
        for (var i = 0; i < n; i++) {
          freq[arr[i]]++;
        }
 
        // Initialise cnt to 0
        var cnt = 0;
 
        // Traversing freq[]
        for (var i = 0; i <= maxE; i++) {
          // If freq of an element is
          // greater than 0 update the
          // value of arr[] at index cnt
          // & increment cnt
          if (freq[i] > 0) {
            var value = 100000 - i;
            arr[cnt] = 100000 * freq[i] + value;
            cnt++;
          }
        }
 
        // Return cnt
        return cnt;
      }
 
      // Function that print array arr[]
      // elements in sorted order
      function printSortedArray(arr, cnt) {
        // Traversing arr[] till index cnt
        for (var i = 0; i < cnt; i++) {
          // Find frequency of elements
          var frequency = parseInt(arr[i] / 100000);
 
          // Find value at index i
          var value = 100000 - (arr[i] % 100000);
 
          // Traversing till frequency
          // to print value at index i
          for (var j = 0; j < frequency; j++) {
            document.write(value + " ");
          }
        }
      }
 
      // Driver code
      var arr = [4, 4, 5, 6, 4, 2, 2, 8, 5];
 
      // Size of array arr[]
      var n = arr.length;
 
      // Function call to get cnt
      var cnt = sortByFreq(arr, n);
 
      // Sort the arr[] in decreasing order
      arr.sort((a, b) => b - a);
 
      // Function that prints elements
      // in decreasing order
      printSortedArray(arr, cnt);
       
</script>

Output: 

4 4 4 2 2 5 5 6 8

 

Time Complexity: O(N*log N) 
Auxiliary Space: O(N)

STL Pair and Comparator based approach :

Approach:

1. Store the frequency of each element in a map.

2. Iterate the map and store the each element and it’s frequency in a vector of pairs.

3. Pass a comparator which sorts the elements in decreasing order of their frequency and by elements value if frequency is equal.

4.Push the elements their frequency number of times inside the final array.

C++




#include<bits/stdc++.h>
using namespace std;
bool compare(pair<int,int>p1,pair<int,int>p2)
{
    if(p1.second==p2.second) //if frequency is equal
    {
        return p1.first<p2.first; //return the smaller value
    }
    return p1.second>p2.second; //return the element with greater frequency
}
int main()
{
    vector<int>arr={4,4,5,6,4,2,2,8,5};
    int n=arr.size();
    map<int,int>m;
    for(int i=0;i<n;i++)
    {
        m[arr[i]]+=1;
    }
    vector<pair<int,int>>a;
    for(auto it=m.begin();it!=m.end();it++)
    {
        a.push_back(make_pair(it->first,it->second)); //making pair of element and it's frequency
    }
    sort(a.begin(),a.end(),compare);
    vector<int>ans;
    for(auto x:a)
    {
        while(x.second--)
        {
        ans.push_back(x.first);
        }
    }
    for(auto x:ans)
    {
        cout<<x<<" ";
    }
    return 0;
}

Python3




# Python code for the approach
from functools import cmp_to_key
 
def compare(p1, p2):
 
    if(p1[1] == p2[1]): #if frequency is equal
        return p1[0] - p2[0] #return the smaller value
     
    return p2[1] - p1[1] #return the element with greater frequency
 
# driver code
arr = [4, 4, 5, 6, 4, 2, 2, 8, 5]
n = len(arr)
m = {}
for i in range(n):
 
    if(arr[i] in m):
        m[arr[i]] = m[arr[i]] + 1
 
    else:
        m[arr[i]] = 1
 
a = []
for x,y in m.items():
    a.append([x,y]) #making pair of element and it's frequency
 
a.sort(key = cmp_to_key(compare))
ans = []
for x in a:
    while(x[1]):
        ans.append(x[0])
        x[1] -= 1
 
for x in ans:
    print(x, end = " ")
 
# This code is contributed by shinjanpatra

Javascript




<script>
 
// JavaScript code for the approach
function compare(p1, p2)
{
    if(p1[1] == p2[1]) //if frequency is equal
    {
        return p1[0] - p2[0]; //return the smaller value
    }
    return p2[1] - p1[1]; //return the element with greater frequency
}
 
// driver code
let arr = [4,4,5,6,4,2,2,8,5];
let n = arr.length;
let m = new Map();
for(let i = 0; i < n; i++)
{
    if(m.has(arr[i])){
        m.set(arr[i], m.get(arr[i]) + 1);
    }
    else m.set(arr[i],1);
}
let a = [];
for(let [x,y] of m)
{
    a.push([x,y]); //making pair of element and it's frequency
}
a.sort(compare);
let ans = [];
for(let x of a){
    while(x[1]--){
        ans.push(x[0]);
    }
}
for(let x of ans)
{
    document.write(x," ");
}
 
// This code is contributed by shinjanpatra
 
</script>

Output

4 4 4 2 2 5 5 6 8 

 


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