# Find duplicates in a given array when elements are not limited to a range

Given an array of n integers. The task is to print the duplicates in the given array. If there are no duplicates then print -1.

Examples:

```Input : {2, 10, 100, 2, 10, 11}
Output : 2 10

Input : {5, 40, 1, 40, 100000, 1, 5, 1}
Output : 5 40 1
```

Note:The duplicate elements can be printed in any order.

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

Simple Approach: By using two loops. It has a time complexity of O(n2).

Efficient Approach: Use unordered_map for hashing. Count frequency of occurrence of each element and the elements with frequency more than 1 is printed. unordered_map is used as range of integers is not known. For Python, Use Dictionary to store number as key and it’s frequency as value. Dictionary can be used as range of integers is not known.

## C++

 `// C++ program to find ` `// duplicates in the given array ` `#include ` `using` `namespace` `std; ` ` `  `// function to find and print duplicates ` `void` `printDuplicates(``int` `arr[], ``int` `n) ` `{ ` `    ``// unordered_map to store frequencies ` `    ``unordered_map<``int``, ``int``> freq; ` `    ``for` `(``int` `i=0; i:: iterator itr; ` `    ``for` `(itr=freq.begin(); itr!=freq.end(); itr++) ` `    ``{ ` `        ``// if frequency is more than 1 ` `        ``// print the element ` `        ``if` `(itr->second > 1) ` `        ``{ ` `            ``cout << itr->first << ``" "``; ` `            ``dup = ``true``; ` `        ``} ` `    ``} ` ` `  `    ``// no duplicates present ` `    ``if` `(dup == ``false``) ` `        ``cout << ``"-1"``; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `arr[] = {12, 11, 40, 12, 5, 6, 5, 12, 11}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``printDuplicates(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find ` `// duplicates in the given array ` `import` `java.util.HashMap; ` `import` `java.util.Map; ` `import` `java.util.Map.Entry; ` ` `  `public` `class` `FindDuplicatedInArray ` `{ ` `    ``// Driver program ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = {``12``, ``11``, ``40``, ``12``, ``5``, ``6``, ``5``, ``12``, ``11``}; ` `        ``int` `n = arr.length; ` `        ``printDuplicates(arr, n); ` `    ``} ` `    ``// function to find and print duplicates ` `    ``private` `static` `void` `printDuplicates(``int``[] arr, ``int` `n)  ` `    ``{ ` `        ``Map map = ``new` `HashMap<>(); ` `        ``int` `count = ``0``; ` `        ``boolean` `dup = ``false``; ` `        ``for``(``int` `i = ``0``; i < n; i++){ ` `            ``if``(map.containsKey(arr[i])){ ` `                ``count = map.get(arr[i]); ` `                ``map.put(arr[i], count + ``1``); ` `            ``} ` `            ``else``{ ` `                ``map.put(arr[i], ``1``); ` `            ``} ` `        ``} ` `         `  `        ``for``(Entry entry : map.entrySet()) ` `        ``{ ` `            ``// if frequency is more than 1 ` `            ``// print the element ` `            ``if``(entry.getValue() > ``1``){ ` `                ``System.out.print(entry.getKey()+ ``" "``); ` `                ``dup = ``true``; ` `            ``} ` `        ``} ` `        ``// no duplicates present ` `        ``if``(!dup){ ` `            ``System.out.println(``"-1"``); ` `        ``} ` `    ``} ` `} `

## Python3

 `# Python 3 code to find duplicates  ` `# using dictionary approach. ` `def` `printDuplicates(arr): ` `    ``dict` `=` `{} ` ` `  `    ``for` `ele ``in` `arr: ` `        ``try``: ` `            ``dict``[ele] ``+``=` `1` `        ``except``: ` `            ``dict``[ele] ``=` `1` `             `  `    ``for` `item ``in` `dict``: ` `         `  `         ``# if frequency is more than 1 ` `         ``# print the element ` `        ``if``(``dict``[item] > ``1``): ` `            ``print``(item, end``=``" "``) ` ` `  `    ``print``(````" "````) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``list` `=` `[``12``, ``11``, ``40``, ``12``,  ` `            ``5``, ``6``, ``5``, ``12``, ``11``] ` `    ``printDuplicates(``list``) ` ` `  `# This code is contributed ` `# by Sushil Bhile `

Output:

```12 11 5
```

Time Complexity: O(n)

## tags:

Arrays Hash Arrays Hash