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Print All Distinct Elements of a given integer array

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  • Difficulty Level : Easy
  • Last Updated : 27 Aug, 2022
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Given an integer array, print all distinct elements in array. The given array may contain duplicates and the output should print every element only once. The given array is not sorted.

Examples: 

Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 12, 10, 9, 45, 2

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1, 2, 3, 4, 5

Input: arr[] = {1, 1, 1, 1, 1}
Output: 1

A Simple Solution is to use two nested loops. The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the element is present on left side of it. If present, then ignores the element, else prints the element. Following is the implementation of the simple algorithm. 

Implementation:

C++




// C++ program to print all distinct elements in a given array
#include <bits/stdc++.h>
using namespace std;
 
void printDistinct(int arr[], int n)
{
    // Pick all elements one by one
    for (int i=0; i<n; i++)
    {
        // Check if the picked element is already printed
        int j;
        for (j=0; j<i; j++)
           if (arr[i] == arr[j])
               break;
 
        // If not printed earlier, then print it
        if (i == j)
          cout << arr[i] << " ";
    }
}
 
// Driver program to test above function
int main()
{
    int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
    int n = sizeof(arr)/sizeof(arr[0]);
    printDistinct(arr, n);
    return 0;
}

Java




// Java program to print all distinct
// elements in a given array
import java.io.*;
 
class GFG {
 
    static void printDistinct(int arr[], int n)
    {
        // Pick all elements one by one
        for (int i = 0; i < n; i++)
        {
            // Check if the picked element
            // is already printed
            int j;
            for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
     
            // If not printed earlier,
            // then print it
            if (i == j)
            System.out.print( arr[i] + " ");
        }
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.length;
        printDistinct(arr, n);
 
    }
}
 
// This code is contributed by vt_m

Python3




# python program to print all distinct
# elements in a given array
 
def printDistinct(arr, n):
 
    # Pick all elements one by one
    for i in range(0, n):
 
        # Check if the picked element
        # is already printed
        d = 0
        for j in range(0, i):
            if (arr[i] == arr[j]):
                d = 1
                break
 
        # If not printed earlier,
        # then print it
        if (d == 0):
            print(arr[i])
     
# Driver program to test above function
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10]
n = len(arr)
printDistinct(arr, n)
 
# This code is contributed by Sam007.

C#




// C# program to print all distinct
// elements in a given array
using System;
 
class GFG {
 
    static void printDistinct(int []arr, int n)
    {
         
        // Pick all elements one by one
        for (int i = 0; i < n; i++)
        {
             
            // Check if the picked element
            // is already printed
            int j;
            for (j = 0; j < i; j++)
                if (arr[i] == arr[j])
                     break;
     
            // If not printed earlier,
            // then print it
            if (i == j)
            Console.Write(arr[i] + " ");
        }
    }
     
    // Driver program
    public static void Main ()
    {
        int []arr = {6, 10, 5, 4, 9, 120,
                                  4, 6, 10};
        int n = arr.Length;
         
        printDistinct(arr, n);
 
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP program to print all distinct
// elements in a given array
 
function printDistinct($arr, $n)
{
    // Pick all elements one by one
    for($i = 0; $i < $n; $i++)
    {
         
        // Check if the picked element
        // is already printed
        $j;
        for($j = 0; $j < $i; $j++)
        if ($arr[$i] == $arr[$j])
            break;
 
        // If not printed
        // earlier, then print it
        if ($i == $j)
        echo $arr[$i] , " ";
    }
}
 
    // Driver Code
    $arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10);
    $n = sizeof($arr);
    printDistinct($arr, $n);
     
// This code is contributed by nitin mittal
?>

Javascript




<script>
//Program to print all distinct elements in a given array
 
   
    function  printDistinct(arr, n)
{
    // Pick all elements one by one
    for (let i=0; i<n; i++)
    {
        // Check if the picked element is already printed
        var j;
        for (j=0; j<i; j++)
           if (arr[i] == arr[j])
               break;
   
        // If not printed earlier, then print it
        if (i == j)
          document.write(arr[i] + " ");
    }
}
   
// Driver program to test above function
    arr = new Array(6, 10, 5, 4, 9, 120, 4, 6, 10);
    n = arr.length;
    printDistinct(arr, n);
//This code is contributed by simranarora5sos
</script>

Output

6 10 5 4 9 120 

Time Complexity: O(n2). 
Auxiliary Space: O(1), since no extra space has been taken.

We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, first sort the array so that all occurrences of every element become consecutive. Once the occurrences become consecutive, we can traverse the sorted array and print distinct elements in O(n) time. Following is the implementation of the idea. 

Implementation:

C++




// C++ program to print all distinct elements in a given array
#include <bits/stdc++.h>
using namespace std;
 
void printDistinct(int arr[], int n)
{
    // First sort the array so that all occurrences become consecutive
    sort(arr, arr + n);
 
    // Traverse the sorted array
    for (int i=0; i<n; i++)
    {
       // Move the index ahead while there are duplicates
       while (i < n-1 && arr[i] == arr[i+1])
          i++;
 
       // print last occurrence of the current element
       cout << arr[i] << " ";
    }
}
 
// Driver program to test above function
int main()
{
    int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
    int n = sizeof(arr)/sizeof(arr[0]);
    printDistinct(arr, n);
    return 0;
}

Java




// Java program to print all distinct
// elements in a given array
import java.io.*;
import java .util.*;
 
class GFG
{
    static void printDistinct(int arr[], int n)
    {
        // First sort the array so that
        // all occurrences become consecutive
        Arrays.sort(arr);
     
        // Traverse the sorted array
        for (int i = 0; i < n; i++)
        {
            // Move the index ahead while
            // there are duplicates
            while (i < n - 1 && arr[i] == arr[i + 1])
                i++;
     
            // print last occurrence of
            // the current element
            System.out.print(arr[i] +" ");
        }
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.length;
        printDistinct(arr, n);
 
    }
}
 
// This code is contributed by vt_m

Python3




# Python program to print all distinct
# elements in a given array
 
def printDistinct(arr, n):
     
    # First sort the array so that
    # all occurrences become consecutive
    arr.sort();
 
    # Traverse the sorted array
    for i in range(n):
         
        # Move the index ahead while there are duplicates
        if(i < n-1 and arr[i] == arr[i+1]):
            while (i < n-1 and (arr[i] == arr[i+1])):
                i+=1;
             
 
        # print last occurrence of the current element
        else:
            print(arr[i], end=" ");
 
# Driver code
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10];
n = len(arr);
printDistinct(arr, n);
 
# This code has been contributed by 29AjayKumar

C#




// C# program to print all distinct
// elements in a given array
using System;
 
class GFG {
 
    static void printDistinct(int []arr, int n)
    {
         
        // First sort the array so that
        // all occurrences become consecutive
        Array.Sort(arr);
     
        // Traverse the sorted array
        for (int i = 0; i < n; i++)
        {
             
            // Move the index ahead while
            // there are duplicates
            while (i < n - 1 && arr[i] == arr[i + 1])
                i++;
     
            // print last occurrence of
            // the current element
            Console.Write(arr[i] + " ");
        }
    }
     
    // Driver program
    public static void Main ()
    {
        int []arr = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.Length;
         
        printDistinct(arr, n);
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP program to print all distinct
// elements in a given array
 
function printDistinct( $arr, $n)
{
     
    // First sort the array so
    // that all occurrences
    // become consecutive
    sort($arr);
 
    // Traverse the sorted array
    for ($i = 0; $i < $n; $i++)
    {
         
        // Move the index ahead
        // while there are duplicates
        while ($i < $n - 1 and
               $arr[$i] == $arr[$i + 1])
            $i++;
     
        // print last occurrence
        // of the current element
        echo $arr[$i] , " ";
    }
}
 
    // Driver Code
    $arr = array(6, 10, 5, 4, 9, 120, 4, 6, 10);
    $n = count($arr);
    printDistinct($arr, $n);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// JavaScript program to print all
// distinct elements in a given array
 
function printDistinct(arr, n)
{
    // First sort the array so that all
    // occurrences become consecutive
    arr.sort((a, b) => a - b);
 
    // Traverse the sorted array
    for (let i=0; i<n; i++)
    {
    // Move the index ahead while
    // there are duplicates
    while (i < n-1 && arr[i] == arr[i+1])
        i++;
 
    // print last occurrence of the
    // current element
    document.write(arr[i] + " ");
    }
}
 
// Driver program to test above function
    let arr = [6, 10, 5, 4, 9, 120, 4, 6, 10];
    let n = arr.length;
    printDistinct(arr, n);
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output

4 5 6 9 10 120 

Time Complexity: O(n log n).
Auxiliary Space: O(1)

We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from left to right and keep track of visited elements in a hash table. Following is the implementation of the idea.

Implementation:

C++




/* CPP program to print all distinct elements
   of a given array */
#include<bits/stdc++.h>
using namespace std;
 
// This function prints all distinct elements
void printDistinct(int arr[],int n)
{
    // Creates an empty hashset
    unordered_set<int> s;
 
    // Traverse the input array
    for (int i=0; i<n; i++)
    {
        // if element is not present then s.count(element) return 0 else return 1
        // hashtable and print it
        if (!s.count(arr[i]))  // <--- avg O(1) time
        {
            s.insert(arr[i]);
            cout << arr[i] << " ";
        }
    }
}
 
// Driver method to test above method
int main ()
{
    int arr[] = {10, 5, 3, 4, 3, 5, 6};
    int n=7;
    printDistinct(arr,n);
    return 0;
}

Java




/* Java program to print all distinct elements of a given array */
import java.util.*;
 
class Main
{
    // This function prints all distinct elements
    static void printDistinct(int arr[])
    {
        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();
 
        // Traverse the input array
        for (int i=0; i<arr.length; i++)
        {
            // If not present, then put it in hashtable and print it
            if (!set.contains(arr[i]))
            {
                set.add(arr[i]);
                System.out.print(arr[i] + " ");
            }
        }
    }
 
    // Driver method to test above method
    public static void main (String[] args)
    {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        printDistinct(arr);
    }
}

Python3




# Python3 program to print all distinct elements
# of a given array
 
# This function prints all distinct elements
def printDistinct(arr, n):
     
    # Creates an empty hashset
    s = dict();
 
    # Traverse the input array
    for i in range(n):
         
        # If not present, then put it in
        # hashtable and print it
        if (arr[i] not in s.keys()):
            s[arr[i]] = arr[i];
            print(arr[i], end = " ");
  
# Driver Code
arr = [10, 5, 3, 4, 3, 5, 6];
n = 7;
printDistinct(arr, n);
 
# This code is contributed by Princi Singh

C#




// C# program to print all distinct
// elements of a given array
using System;
using System.Collections.Generic;
 
class GFG
{
// This function prints all
// distinct elements
public static void printDistinct(int[] arr)
{
    // Creates an empty hashset
    HashSet<int> set = new HashSet<int>();
 
    // Traverse the input array
    for (int i = 0; i < arr.Length; i++)
    {
        // If not present, then put it
        // in hashtable and print it
        if (!set.Contains(arr[i]))
        {
            set.Add(arr[i]);
            Console.Write(arr[i] + " ");
        }
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = new int[] {10, 5, 3, 4, 3, 5, 6};
    printDistinct(arr);
}
}
 
// This code is contributed by Shrikant13

Javascript




<script>
 
// Javascript program to print all distinct elements of a given array
 
    // This function prints all distinct elements
    function printDistinct(arr)
    {
        // Creates an empty hashset
        let set = new Set();
  
        // Traverse the input array
        for (let i=0; i<arr.length; i++)
        {
            // If not present, then put it in hashtable and print it
            if (!set.has(arr[i]))
            {
                set.add(arr[i]);
                document.write(arr[i] + " ");
            }
        }
    }
 
// Driver program
 
      let arr = [10, 5, 3, 4, 3, 5, 6];
        printDistinct(arr);
       
</script>

Output

10 5 3 4 6 

Time Complexity: O(n).
Auxiliary Space: O(n)

One more advantage of hashing over sorting is, the elements are printed in same order as they are in input array. 

Another Approach: 

  1.  Put all input integers to hashmap’s key 
  2. Print keySet outside the loop 

Implementation:

C++




// C++ approach
#include <bits/stdc++.h>
using namespace std;
 
int main() {
  int ar[] = { 10, 5, 3, 4, 3, 5, 6 };
  map<int  ,int> hm;
  for (int i = 0; i < sizeof(ar)/sizeof(ar[0]); i++) { // total = O(n*logn)
    hm.insert({ar[i], i});  //  time complexity for insert() in map O(logn)
  }
  cout <<"[";
  for (auto const &pair: hm) {
    cout << pair.first << ", ";
  }
  cout <<"]";
}
 
// This code is contributed by Shubham Singh

Java




import java.util.HashMap;
public class UniqueInArray2 {
 
    public static void main(String args[])
 
    {
        int ar[] = { 10, 5, 3, 4, 3, 5, 6 };
        HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();
        for (int i = 0; i < ar.length; i++) {
            hm.put(ar[i], i);
        }
        // Using hm.keySet() to print output
        // reduces time complexity. - Lokesh
        System.out.println(hm.keySet());
 
    }
 
}

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
public class UniqueInArray2
{
 
    public static void Main(String []args)
 
    {
        int []ar = { 10, 5, 3, 4, 3, 5, 6 };
        Dictionary<int,int> hm = new Dictionary<int,int>();
        for (int i = 0; i < ar.Length; i++)
        {
            if(hm.ContainsKey(ar[i]))
                hm.Remove(ar[i]);
            hm.Add(ar[i], i);
        }
 
        // Using hm.Keys to print output
        // reduces time complexity. - Lokesh
        var v = hm.Keys;
        foreach(int a in v)
            Console.Write(a+" ");
 
    }
 
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
        var ar = [ 10, 5, 3, 4, 3, 5, 6 ];
        var hm = new Map();
        for (i = 0; i < ar.length; i++) {
            hm.set(ar[i], i);
        }
         
        // Using hm.keySet() to print output
        // reduces time complexity. - Lokesh
        var key = hm.keys();
        for(var k of key)
        document.write(k+" ");
 
// This code is contributed by gauravrajput1
</script>

Python3




ar = [ 10, 5, 3, 4, 3, 5, 6 ];
hm = {};
for i in range(len(ar)):
    hm[ar[i]] = i;
 
# Using hm.keySet() to print output
# reduces time complexity. - Lokesh
print(hm.keys());
 
# This code contributed by shikhasingrajput

Output

[3, 4, 5, 6, 10, ]

Time Complexity: O(n*logn)
Auxiliary Space: O(n)


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