Find the frequencies of all duplicates elements in the array

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)