Encuentre las frecuencias de todos los elementos duplicados en la array

Dada una array de enteros con elementos duplicados, la tarea es encontrar los elementos duplicados en la array y sus frecuencias.
Ejemplos: 
 

Entrada: arr[] = {2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6} 
Salida: A continuación se muestra la frecuencia de los elementos repetidos – 
4 –> 4 
5 –> 2 
6 –> 3
Entrada: arr[] = {4, 4, 5, 5, 6} 
Salida: A continuación se muestra la frecuencia de los elementos repetidos – 
4 –> 2 
5 –> 2 
 

Acercarse: 
 

• Cree un Hash Map para almacenar la frecuencia de los elementos.
• Los elementos cuya frecuencia es mayor que 1 son los elementos repetidos.

A continuación se muestra la implementación del enfoque anterior: 
 

// CPP Implementation to find the
// repeating elements with there count
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the repeating
// elements with there count
map<int,int> findRepeating(int arr[], int size){
 
    // Hash map to store the
    // frequency of elements
    map<int,int> frequency;
     
    // Loop to store the frequency of
    // elements of array
    for (int i = 0; i < size; i++)
    frequency[arr[i]]++;
    return frequency;
}
         
// Driver Code
int main(){
    int arr[] = {4, 4, 5, 5, 6};
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    map<int,int> frequency = findRepeating(arr, arr_size);
    cout<<"Below is the frequency of repeated elements -"<<endl;
    for(auto x : frequency){
        if (frequency[x.first] > 1)
            cout<<x.first<<" --> "<<frequency[x.first]<<endl;
    }
}
     
// This code is contributed by Surendra_Gangwar

// Java Implementation to find the
// repeating elements with there count
import java.util.*;
 
class GFG
{
     
// Function to find the repeating
// elements with there count
static HashMap<Integer, Integer> findRepeating(int []arr, int size){
     
    // Hash map to store the
    // frequency of elements
    HashMap<Integer,Integer> frequency = new HashMap<Integer,Integer>();
     
    // Loop to store the frequency of
    // elements of array
    for(int i = 0; i < size; i++)
    {
        if(frequency.containsKey(arr[i]))
        {
            frequency.put(arr[i], frequency.get(arr[i]) + 1);
        }
        else
        {
            frequency.put(arr[i], 1);
        }
    }
    return frequency;
}
 
// Driver Code
public static void main(String []args)
{
    int []arr = {4, 4, 5, 5, 6};
    int arr_size = arr.length;
    HashMap<Integer,Integer> frequency = findRepeating(arr, arr_size);
    System.out.println("Below is the frequency"
    +"of repeated elements -");
    for (Map.Entry<Integer,Integer> entry : frequency.entrySet())
        if (entry.getValue() > 1)
            System.out.println(entry.getKey()+ " --> "+entry.getValue());
}
}
 
// This code is contributed by PrinciRaj1992

# Python Implementation to find the
# repeating elements with there count
 
# Function to find the repeating
# elements with there count
def findRepeating(arr, size):
     
    # Hash map to store the
    # frequency of elements
    frequency = {}
     
    # Loop to store the frequency of
    # elements of array
    for i in range (0, size):
        frequency[arr[i]] = \
        frequency.get(arr[i], 0) + 1
    return frequency
     
# Driver Code
if __name__ == "__main__":
    arr = [4, 4, 5, 5, 6]
    arr_size = len(arr)
    frequency = findRepeating(arr, arr_size)
    print("Below is the frequency\
    of repeated elements -")
    for i in frequency:
        if frequency[i] > 1:
            print(i, " --> ", frequency[i])

// C# Implementation to find the
// repeating elements with there count
using System;
using System.Collections.Generic;
 
class GFG
{
     
// Function to find the repeating
// elements with there count
static Dictionary<int, int> findRepeating(int []arr, int size){
     
    // Hash map to store the
    // frequency of elements
    Dictionary<int,int> frequency = new Dictionary<int,int>();
     
    // Loop to store the frequency of
    // elements of array
    for(int i = 0; i < size; i++)
    {
        if(frequency.ContainsKey(arr[i]))
        {
            frequency[arr[i]] = frequency[arr[i]] + 1;
        }
        else
        {
            frequency.Add(arr[i], 1);
        }
    }
    return frequency;
}
 
// Driver Code
public static void Main(String []args)
{
    int []arr = {4, 4, 5, 5, 6};
    int arr_size = arr.Length;
    Dictionary<int,int> frequency = findRepeating(arr, arr_size);
    Console.WriteLine("Below is the frequency"
                        +"of repeated elements -");
    foreach (KeyValuePair<int,int> entry in frequency)
        if (entry.Value > 1)
            Console.WriteLine(entry.Key+ " --> "+entry.Value);
}
}
 
// This code is contributed by 29AjayKumar

<script>
// Javascript Implementation to find the
// repeating elements with there count
     
// Function to find the repeating
// elements with there count
function findRepeating(arr, size)
{
     
    // Hash map to store the
    // frequency of elements
    var frequency = new Map();
     
    // Loop to store the frequency of
    // elements of array
    for(var i = 0; i < size; i++)
    {
        if(frequency.has(arr[i]))
        {
            frequency.set(arr[i], frequency.get(arr[i])+1);
        }
        else
        {
            frequency.set(arr[i], 1);
        }
    }
    return frequency;
}
 
// Driver Code
var arr = [4, 4, 5, 5, 6];
var arr_size = arr.length;
var frequency = findRepeating(arr, arr_size);
document.write("Below is the frequency"
                    +"of repeated elements -<br>");
frequency.forEach((value, key) => {
    if (value > 1)
        document.write(key+ " --> "+value + "<br>");
});
 
// This code is contributed by rrrtnx.
</script>

Below is the frequency    of repeated elements -
4  -->  2
5  -->  2