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Find the frequencies of all duplicates elements in the array

Last Updated : 21 Mar, 2023
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Given an array of integers with duplicate elements in it, the task is to find the duplicate elements in the array and their frequencies.

Examples: 

Input: arr[] = {2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6} 
Output: Below is the frequency of repeated elements – 
4 –> 4 
5 –> 2 
6 –> 3
Input: arr[] = {4, 4, 5, 5, 6} 
Output: Below is the frequency of repeated elements – 
4 –> 2 
5 –> 2 

Approach: 

  • Create a Hash Map to store the frequency of the elements.
  • Elements whose frequency is greater than 1 are the repeated elements.

Below is the implementation of the above approach: 

CPP




// 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




// 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




# 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#




// 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


Javascript




<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>


Output: 

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

 

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

Efficient Approach( Space optimization): we can use binary search . For this , First sort the array and then find frequency of all array element with the use of binary search function ( Upper_bound ) . The frequency of array element will be ‘last_index-first_index+1’ . If the frequency is greater than one , then print it .      

Below is the implementation of the above approach: 

C++




// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
//Function to find frequency of elements in the array
void findRepeating(int arr[], int n)
{  
    sort(arr,arr+n);//sort array for binary search
    
    for(int i = 0 ; i < n ;i++)
    {
      //index of first and last occ of arr[i]
      int first_index = lower_bound(arr,arr+n,arr[i])- arr;
      int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
      i=last_index;
       
      int fre = last_index-first_index+1;//finding frequency
      if(fre >= 2)
      { //printing frequency if it is greater than 1
        cout << arr[i] <<" --> "<<fre <<endl;
      }
    }
}
 
// Drive code
int main()
{  
    int arr[] = {2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
   
    // Function call
    findRepeating(arr, n);
    return 0;
     
}
 
// This Code is contributed by nikhilsainiofficial546


C#




using System;
 
class MainClass {
    // Function to find frequency of elements in the array
    static void FindRepeating(int[] arr, int n)
    {
        // sort array for binary search
        Array.Sort(arr);
 
        for (int i = 0; i < n; i++) {
            // index of first and last occ of arr[i]
            int first_index = Array.IndexOf(arr, arr[i]);
            int last_index = Array.LastIndexOf(arr, arr[i]);
            i = last_index;
 
            // finding frequency
            int fre = last_index - first_index + 1;
            if (fre >= 2) {
                // printing frequency if it is greater than
                // 1
                Console.WriteLine(arr[i] + " --> " + fre);
            }
        }
    }
 
    // Drive code
    public static void Main(string[] args)
    {
        int[] arr = { 2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6 };
        int n = arr.Length;
 
        // Function call
        FindRepeating(arr, n);
    }
}


Java




import java.util.*;
 
class Main {
    // Function to find frequency of elements in the array
    public static void findRepeating(int arr[], int n)
    {
        Arrays.sort(arr); // sort array for binary search
 
        int i = 0;
        while (i < n) {
            // index of first and last occ of arr[i]
            int firstIndex = i;
            while (i < n - 1 && arr[i] == arr[i + 1]) {
                i++;
            }
            int lastIndex = i;
 
            int frequency = lastIndex - firstIndex
                            + 1; // finding frequency
            if (frequency >= 2) {
                // printing frequency if it is greater than
                // 1
                System.out.println(arr[i] + " --> "
                                   + frequency);
            }
            i++;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6 };
        int n = arr.length;
 
        // Function call
        findRepeating(arr, n);
    }
}


Javascript




//Function to find frequency of elements in the array
function findRepeating(arr, n) {  
    arr.sort(); // sort array for binary search
 
    for(let i = 0 ; i < n ; i++) {
        // index of first and last occ of arr[i]
        let first_index = arr.indexOf(arr[i]);
        let last_index = arr.lastIndexOf(arr[i]);
        i = last_index;
         
        let fre = last_index - first_index + 1; // finding frequency
        if(fre >= 2) { // printing frequency if it is greater than 1
            console.log(arr[i] + " --> " + fre);
        }
    }
}
 
// Drive code
let arr = [2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6];
let n = arr.length;
 
// Function call
findRepeating(arr, n);


Python3




def find_repeating(arr, n):
    arr.sort() # sort array for binary search
 
    i = 0
    while i < n:
        # index of first and last occ of arr[i]
        first_index = i
        while i < n - 1 and arr[i] == arr[i + 1]:
            i += 1
        last_index = i
 
        frequency = last_index - first_index + 1 # finding frequency
        if frequency >= 2:
            # printing frequency if it is greater than 1
            print(str(arr[i]) + " --> " + str(frequency))
        i += 1
 
# Driver code
arr = [2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6]
n = len(arr)
 
# Function call
find_repeating(arr, n)


Output

4 --> 4
5 --> 2
6 --> 3

Time Complexity: O(n*log2n) , Taking O(log2n) time for binary search function ( upper_bound) .
Auxiliary Space: O(1)



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