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[0]);
    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<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

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

<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 Code
let arr=[1, 2, 5, 1, 7, 2, 4, 2 ];
let n = arr.length;
removeDups(arr, n);


// This code is contributed by unknown2108

</script>

Output

1 2 5 7 4 

Complexity Analysis:

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