Skip to content
Related Articles

Related Articles

Replace every elements in the array by its frequency in the array
  • Difficulty Level : Basic
  • Last Updated : 09 Apr, 2019

Given an array of integers, replace every element by its frequency in the array.

Examples:

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

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

Approach:

  1. Take a hash map, which will store the frequency of all the elements in the array.
  2. Now, traverse once again.
  3. Now, replace all the elements by its frequency.
  4. Print the modified array.

C++




// C++ program to replace the elements
// by their frequency in the array.
#include "iostream"
#include "unordered_map"
using namespace std;
  
void ReplaceElementsByFrequency(int arr[], int n)
{
    // Hash map which will store the
    // frequency of the elements of the array.
    unordered_map<int, int> mp;
  
    for (int i = 0; i < n; ++i) {
  
        // Increment the frequency
        // of the element by 1.
        mp[arr[i]]++;
    }
  
    // Replace every element by its frequency
    for (int i = 0; i < n; ++i) {
        arr[i] = mp[arr[i]];
    }
}
  
int main()
{
    int arr[] = { 1, 2, 5, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    ReplaceElementsByFrequency(arr, n);
  
    // Print the modified array.
    for (int i = 0; i < n; ++i) {
        cout << arr[i] << " ";
    }
    return 0;
}

Java




import java.util.HashMap;
  
// Java program to replace the elements 
// by their frequency in the array.
class GFG {
  
    static void ReplaceElementsByFrequency(int arr[], int n) {
        // Hash map which will store the 
        // frequency of the elements of the array. 
        HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>();
  
        for (int i = 0; i < n; ++i) {
  
            // Increment the frequency 
            // of the element by 1. 
            if (mp.get(arr[i]) == null) {
                mp.put(arr[i], 1);
            } else {
                mp.put(arr[i], (mp.get(arr[i]) + 1));
            }
            //mp[arr[i]]++; 
        }
  
        // Replace every element by its frequency 
        for (int i = 0; i < n; ++i) {
            if (mp.get(arr[i]) != null) {
                arr[i] = mp.get(arr[i]);
            }
            //arr[i] = mp[arr[i]]; 
        }
    }
  
    public static void main(String[] args) {
        int arr[] = {1, 2, 5, 2, 2, 5};
        int n = arr.length;
  
        ReplaceElementsByFrequency(arr, n);
  
        // Print the modified array. 
        for (int i = 0; i < n; ++i) {
            System.out.print(arr[i] + " ");
        }
    }
}
// This code contributed by 29AJayKumar

Python3




# Python 3 program to replace the elements
# by their frequency in the array.
  
def ReplaceElementsByFrequency(arr, n):
      
    # Hash map which will store the
    # frequency of the elements of the array.
    mp = {i:0 for i in range(len(arr))}
  
    for i in range(n):
          
        # Increment the frequency of the 
        # element by 1.
        mp[arr[i]] += 1
  
    # Replace every element by its frequency
    for i in range(n):
        arr[i] = mp[arr[i]]
  
# Driver Code
if __name__ == '__main__':
    arr = [1, 2, 5, 2, 2, 5]
    n = len(arr)
  
    ReplaceElementsByFrequency(arr, n);
  
    # Print the modified array.
    for i in range(n):
        print(arr[i], end = " ")
  
# This code is contributed by
# Sahil_shelangia

C#




// C# program to replace the elements 
// by their frequency in the array.
using System;
using System.Collections.Generic; 
      
class GFG 
{
  
    static void ReplaceElementsByFrequency(int []arr, int n) 
    {
        // Hash map which will store the 
        // frequency of the elements of the array. 
        Dictionary<int,int> mp = new Dictionary<int,int>();
  
        for (int i = 0; i < n; ++i)
        {
  
            // Increment the frequency 
            // of the element by 1. 
            if (!mp.ContainsKey(arr[i])) 
            {
                mp.Add(arr[i], 1);
            
            else
            {
                var a = mp[arr[i]] + 1;
                mp.Remove(arr[i]);
                mp.Add(arr[i], a);
            }
        }
  
        // Replace every element by its frequency 
        for (int i = 0; i < n; ++i) 
        {
            if (mp[arr[i]] != 0) 
            {
                arr[i] = mp[arr[i]];
            }
            //arr[i] = mp[arr[i]]; 
        }
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int []arr = {1, 2, 5, 2, 2, 5};
        int n = arr.Length;
  
        ReplaceElementsByFrequency(arr, n);
  
        // Print the modified array. 
        for (int i = 0; i < n; ++i)
        {
            Console.Write(arr[i] + " ");
        }
    }
}
  
// This code contributed by Rajput-Ji

Output :

1 3 2 3 3 2

Time Complexity – O(N)




My Personal Notes arrow_drop_up
Recommended Articles
Page :