Remove duplicates from unsorted array using Map data structure

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++
Java
Python3
C#
Javascript
// 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[0]);    removeDups(arr, n);    return 0;}
// 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<Integer,                Boolean> 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
# Python 3 program to remove the# duplicates from the arraydef 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 codearr = [ 1, 2, 5, 1, 7, 2, 4, 2 ] # len of arrayn = len(arr) removeDups(arr,n) # This code is contributed# by Mohit Kumar
// 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
<script> // JavaScript program to remove// the duplicates from the array. function removeDups(arr,n){    // Hash map which will store the        // elements which has appeared previously.        let mp = new Map();           for (let i = 0; i < n; ++i)        {               // Print the element if it is not            // there in the hash map            if (mp.get(arr[i]) == null)                document.write(arr[i] + " ");               // Insert the element in the hash map            mp.set(arr[i], true);        }} // Driver Codelet arr=[1, 2, 5, 1, 7, 2, 4, 2 ];let n = arr.length;removeDups(arr, n);  // This code is contributed by unknown2108 </script>
1 2 5 7 4
 
Time Complexity – O(N)
Auxiliary Space – O(N). 
mohit kumar 29
aman2096
sanjeev2552
Rajput-Ji
unknown2108
anandkumarshivam2266
cpp-unordered_map
Technical Scripter 2018
Arrays
Hash
Technical Scripter
Arrays
Hash
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