# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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);  ` `    ``map<``int``,``int``> frequency = findRepeating(arr, arr_size); ` `    ``cout<<``"Below is the frequency of repeated elements -"``< 1) ` `            ``cout< "``<

## 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 findRepeating(``int` `[]arr, ``int` `size){ ` `     `  `    ``// Hash map to store the  ` `    ``// frequency of elements ` `    ``HashMap frequency = ``new` `HashMap(); ` `     `  `    ``// 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 frequency = findRepeating(arr, arr_size); ` `    ``System.out.println(``"Below is the frequency"` `    ``+``"of repeated elements -"``); ` `    ``for` `(Map.Entry 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 `

Output:

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

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