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# Remove duplicates from unsorted array using Map data structure

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.

Implementation:

## 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 `

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(N)

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