# Remove duplicates from unsorted array using Map data structure

• Difficulty Level : Easy
• Last Updated : 31 May, 2022

Given an unsorted array of integers, print the array after removing the duplicate elements from it. We need to print distinct array elements according to their first occurrence.
Examples:

```Input : arr[] = { 1, 2, 5, 1, 7, 2, 4, 2}
Output : 1 2 5 7 4
Explanation : {1, 2} appear more than one time.```

Approach :

• Take a hash map, which will store all the elements which have appeared before.
• Traverse the array.
• Check if the element is present in the hash map.
• If yes, continue traversing the array.
• Else Print the element.

## C++

 `// C++ program to remove the duplicates from the array.``#include "iostream"``#include "unordered_map"``using` `namespace` `std;` `void` `removeDups(``int` `arr[], ``int` `n)``{``    ``// Hash map which will store the``    ``// elements which has appeared previously.``    ``unordered_map<``int``, ``bool``> mp;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Print the element if it is not``        ``// there in the hash map``        ``if` `(mp.find(arr[i]) == mp.end()) {``            ``cout << arr[i] << ``" "``;``        ``}` `        ``// Insert the element in the hash map``        ``mp[arr[i]] = ``true``;``    ``}``}` `int` `main(``int` `argc, ``char` `const``* argv[])``{``    ``int` `arr[] = { 1, 2, 5, 1, 7, 2, 4, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``removeDups(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to remove``// the duplicates from the array.``import` `java.util.HashMap;` `class` `GFG``{``    ``static` `void` `removeDups(``int``[] arr, ``int` `n)``    ``{` `        ``// Hash map which will store the``        ``// elements which has appeared previously.``        ``HashMap mp = ``new` `HashMap<>();` `        ``for` `(``int` `i = ``0``; i < n; ++i)``        ``{` `            ``// Print the element if it is not``            ``// there in the hash map``            ``if` `(mp.get(arr[i]) == ``null``)``                ``System.out.print(arr[i] + ``" "``);` `            ``// Insert the element in the hash map``            ``mp.put(arr[i], ``true``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``1``, ``2``, ``5``, ``1``, ``7``, ``2``, ``4``, ``2` `};``        ``int` `n = arr.length;``        ``removeDups(arr, n);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python 3 program to remove the``# duplicates from the array``def` `removeDups(arr, n):``     ` `    ``# dict to store every element``    ``# one time``    ``mp ``=` `{i : ``0` `for` `i ``in` `arr}``    ` `    ``for` `i ``in` `range``(n):``        ` `        ``if` `mp[arr[i]] ``=``=` `0``:``            ``print``(arr[i], end ``=` `" "``)``            ``mp[arr[i]] ``=` `1` `# Driver code``arr ``=` `[ ``1``, ``2``, ``5``, ``1``, ``7``, ``2``, ``4``, ``2` `]` `# len of array``n ``=` `len``(arr)` `removeDups(arr,n)` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# program to remove``// the duplicates from the array.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``static` `void` `removeDups(``int``[] arr, ``int` `n)``    ``{`` ` `        ``// Hash map which will store the``        ``// elements which has appeared previously.``        ``Dictionary<``int``,``                ``Boolean> mp = ``new` `Dictionary<``int``, Boolean>();`` ` `        ``for` `(``int` `i = 0; i < n; ++i)``        ``{`` ` `            ``// Print the element if it is not``            ``// there in the hash map``            ``if` `(!mp.ContainsKey(arr[i]))``                ``Console.Write(arr[i] + ``" "``);`` ` `            ``// Insert the element in the hash map``            ``mp[arr[i]] = ``true``;``        ``}``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 1, 2, 5, 1, 7, 2, 4, 2 };``        ``int` `n = arr.Length;``        ``removeDups(arr, n);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`1 2 5 7 4`

Time Complexity – O(N)

Auxiliary Space – O(N).

My Personal Notes arrow_drop_up