Given an integer array, print all distinct elements in an 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, 2Input: arr[] = {1, 2, 3, 4, 5}
Output: 1, 2, 3, 4, 5Input: arr[] = {1, 1, 1, 1, 1}
Output: 1
Print all Distinct ( Unique ) Elements in given Array using Nested loop:
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 functionint 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 arrayimport 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 arraydef 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 functionarr = [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 arrayusing 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. |
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> |
PHP
<?php// PHP program to print all distinct// elements in a given arrayfunction 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?> |
Output
6 10 5 4 9 120
Time Complexity: O(n2).
Auxiliary Space: O(1), since no extra space has been taken.
Print all Distinct ( Unique ) Elements in given Array using using Sorting :
We can use Sorting to solve the problem in O(N log N) 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 functionint 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 arrayimport 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 arraydef 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 codearr = [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 arrayusing 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. |
Javascript
<script>// JavaScript program to print all// distinct elements in a given arrayfunction 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> |
PHP
<?php// PHP program to print all distinct// elements in a given arrayfunction 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.?> |
Output
4 5 6 9 10 120
Time Complexity: O(n log n).
Auxiliary Space: O(1)
Print all Distinct ( Unique ) Elements in given Array using Set :
The Easiest Way to do it is to take a vector input then put the vector into a set and just traverse over the set.
C++
#include <iostream>#include<bits/stdc++.h>using namespace std;int main() { vector<int>v{10, 5, 3, 4, 3, 5, 6}; set<int>s(v.begin(),v.end()); cout<<"All the distinct element in given vector in sorted order are: "; for(auto it:s)cout<<it<<" "; cout<<endl; return 0;} |
Java
import java.io.*;import java .util.*;class GFG { public static void main (String[] args) { List<Integer>v = new ArrayList<Integer>(); v.add(10); v.add(5); v.add(3); v.add(4); v.add(3); v.add(5); v.add(6); SortedSet<Integer> s = new TreeSet<Integer>(); for(int i=0; i<v.size(); i++) s.add(v.get(i)); System.out.print("All the distinct element in given vector in sorted order are: "); for (Integer value : s) System.out.print(value+" "); }}// This code is contributed by ratiagrawal. |
Python3
from sortedcontainers import SortedList, SortedSet, SortedDictv=[10, 5, 3, 4, 3, 5, 6];s=SortedSet();for i in range (0,len(v)): s.add(v[i]);print("All the distinct element in given vector in sorted order are: ");for it in s: print(it," "); |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG { public static void Main() { int[] v = {10, 5, 3, 4, 3, 5, 6}; SortedSet<int> s = new SortedSet<int>(); for(int i = 0; i < v.Length; i++) s.Add(v[i]); Console.Write("All the distinct element in given vector in sorted order are: "); foreach (int res in s) Console.Write(res+" "); }}// This code is contributed by poojaagrawal2. |
Javascript
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)// JavaScript Program for the above approachlet v = [10, 5, 3, 4, 3, 5, 6];v.sort(function(a,b){return a-b});let s = new Set(v);document.write("All the distinct element in given vector in sorted order are : ");s.forEach(function(value){ document.write(value + " ");}); |
Output
All the distinct element in given vector in sorted order are: 3 4 5 6 10
Time Complexity: O(N.log N).
Auxiliary Space: O(N), for a set.
Print all Distinct ( Unique ) Elements in given Array using Hashing :
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 elementsvoid 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 methodint 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 elementsdef 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 Codearr = [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 Codepublic 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
