Print all Distinct (Unique) Elements in given Array
Last Updated : 02 Nov, 2024
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Given an integer array arr[], print all distinct elements from this array. The given array may contain duplicates and the output should contain every element only once.
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}
Table of Content
[Naive Approach] Using Nested loops - O(n^2) Time and O(1) Space
The idea 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 ignore the element, else store it in result.
// C++ program to print all distinct elements in an
// array using nested loops
#include <iostream>
#include <vector>
using namespace std;
vector<int> findDistinct(vector<int> &arr) {
vector<int> res;
for (int i = 0; i < arr.size(); i++) {
// Check if this element is included in result
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// Include this element if not included previously
if (i == j)
res.push_back(arr[i]);
}
return res;
}
int main() {
vector<int> arr = {12, 10, 9, 45, 2, 10, 10, 45};
vector<int> res = findDistinct(arr);
for (int ele : res)
cout << ele << " ";
return 0;
}
// C program to print all distinct elements in an
// array using nested loops
#include <stdio.h>
int* findDistinct(int *arr, int size, int *resSize) {
int *res = (int*)malloc(size * sizeof(int));
*resSize = 0;
for (int i = 0; i < size; i++) {
// Check if this element is included in result
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// Include this element if not included previously
if (i == j)
res[(*resSize)++] = arr[i];
}
return res;
}
int main() {
int arr[] = {12, 10, 9, 45, 2, 10, 10, 45};
int size = sizeof(arr) / sizeof(arr[0]);
int resSize;
int *res = findDistinct(arr, size, &resSize);
for (int i = 0; i < resSize; i++)
printf("%d ", res[i]);
free(res);
return 0;
}
// JAva program to print all distinct elements in an
// array using nested loops
import java.util.*;
class GfG {
static ArrayList<Integer> findDistinct(int[] arr) {
ArrayList<Integer> res = new ArrayList<>();
for (int i = 0; i < arr.length; i++) {
// Check if this element is included in result
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// Include this element if not included previously
if (i == j)
res.add(arr[i]);
}
return res;
}
public static void main(String[] args) {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};
ArrayList<Integer> res = findDistinct(arr);
for (int val : res) {
System.out.print(val + " ");
}
}
}
# Python program to print all distinct elements in an
# array using nested loops
def findDistinct(arr):
res = []
for i in range(len(arr)):
# Check if this element is included in result
j = 0
while j < i:
if arr[i] == arr[j]:
break
j += 1
# Include this element if not included previously
if i == j:
res.append(arr[i])
return res
if __name__ == "__main__":
arr = [12, 10, 9, 45, 2, 10, 10, 45]
res = findDistinct(arr)
for val in res:
print(val, end=" ")
// C# program to print all distinct elements in an
// array using nested loops
using System;
using System.Collections.Generic;
class GfG {
static List<int> findDistinct(int[] arr) {
List<int> res = new List<int>();
for (int i = 0; i < arr.Length; i++) {
// Check if this element is included in result
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// Include this element if not included previously
if (i == j)
res.Add(arr[i]);
}
return res;
}
static void Main() {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};
List<int> res = findDistinct(arr);
foreach (int val in res) {
Console.Write(val + " ");
}
}
}
// JavaScript program to print all distinct elements in an
// array using nested loops
function findDistinct(arr) {
let res = [];
for (let i = 0; i < arr.length; i++) {
// Check if this element is included in result
let j;
for (j = 0; j < i; j++)
if (arr[i] === arr[j])
break;
// Include this element if not included previously
if (i === j)
res.push(arr[i]);
}
return res;
}
let arr = [12, 10, 9, 45, 2, 10, 10, 45];
let res = findDistinct(arr);
console.log(res.join(" "));
Output
12 10 9 45 2
[Better Approach] Using Sorting - O(n*logn) Time and O(1) Space
The idea is to 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 by ignoring elements if they are same as the previous element.
// C++ program to print all distinct elements in an
// array using sorting
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> findDistinct(vector<int> &arr) {
vector<int> res;
int n = arr.size();
// First sort the array so that all occurrences
// become consecutive
sort(arr.begin(), arr.end());
for (int i = 0; i < n; i++) {
// Store elements only if they are different
// from previous element
if(i == 0 || arr[i] != arr[i - 1]) {
res.push_back(arr[i]);
}
}
return res;
}
int main() {
vector<int> arr = {12, 10, 9, 45, 2, 10, 10, 45};
vector<int> res = findDistinct(arr);
for(int ele: res) {
cout << ele << " ";
}
return 0;
}
// C program to print all distinct elements in an
// array using sorting
#include <stdio.h>
int compare(const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
void findDistinct(int *arr, int n, int **res, int *resSize) {
// First sort the array so that all occurrences
// become consecutive
qsort(arr, n, sizeof(int), compare);
*res = (int *)malloc(n * sizeof(int));
int index = 0;
for (int i = 0; i < n; i++) {
// Store elements only if they are different
// from previous element
if (i == 0 || arr[i] != arr[i - 1]) {
(*res)[index++] = arr[i];
}
}
*resSize = index;
}
int main() {
int arr[] = {12, 10, 9, 45, 2, 10, 10, 45};
int n = sizeof(arr) / sizeof(arr[0]);
int *res;
int resSize;
findDistinct(arr, n, &res, &resSize);
for (int i = 0; i < resSize; i++) {
printf("%d ", res[i]);
}
return 0;
}
// Java program to print all distinct elements in an
// array using sorting
import java.util.Arrays;
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findDistinct(int[] arr) {
ArrayList<Integer> res = new ArrayList<>();
int n = arr.length;
// First sort the array so that all occurrences
// become consecutive
Arrays.sort(arr);
for (int i = 0; i < n; i++) {
// Store elements only if they are different
// from previous element
if (i == 0 || arr[i] != arr[i - 1]) {
res.add(arr[i]);
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};
ArrayList<Integer> res = findDistinct(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
# Python program to print all distinct elements in an
# array using sorting
def findDistinct(arr):
res = []
n = len(arr)
# First sort the array so that all occurrences
# become consecutive
arr.sort()
for i in range(n):
# Store elements only if they are different
# from previous element
if i == 0 or arr[i] != arr[i - 1]:
res.append(arr[i])
return res
if __name__ == "__main__":
arr = [12, 10, 9, 45, 2, 10, 10, 45]
res = findDistinct(arr)
for ele in res:
print(ele, end=" ")
// C# program to print all distinct elements in an
// array using sorting
using System;
using System.Collections.Generic;
class GfG {
static List<int> findDistinct(List<int> arr) {
List<int> res = new List<int>();
int n = arr.Count;
// First sort the array so that all occurrences
// become consecutive
arr.Sort();
for (int i = 0; i < n; i++) {
// Store elements only if they are different
// from previous element
if (i == 0 || arr[i] != arr[i - 1]) {
res.Add(arr[i]);
}
}
return res;
}
static void Main() {
List<int> arr = new List<int> { 12, 10, 9, 45, 2, 10, 10, 45 };
List<int> res = findDistinct(arr);
foreach (int ele in res) {
Console.Write(ele + " ");
}
}
}
// JavaScript program to print all distinct elements in an
// array using sorting
function findDistinct(arr) {
let res = [];
let n = arr.length;
// First sort the array so that all occurrences
// become consecutive
arr.sort((a, b) => a - b);
for (let i = 0; i < n; i++) {
// Store elements only if they are different
// from previous element
if (i === 0 || arr[i] !== arr[i - 1]) {
res.push(arr[i]);
}
}
return res;
}
const arr = [12, 10, 9, 45, 2, 10, 10, 45];
const res = findDistinct(arr);
console.log(res.join(" "));
Output
2 9 10 12 45
[Expected Approach] Using Hashing - O(n) Time and O(n) Space
We can use Hashing to store distinct element. The idea is to insert all the elements in a hash set and then traverse the hash set to store the distinct elements in the resultant array.
// C++ program to print all distinct elements
// of a given array
#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
// function to return all distinct elements
vector<int> findDistinct(vector<int> &arr) {
// Initialize set with all elements of array
unordered_set<int> st (arr.begin(), arr.end());
// Return the result array by inserting all
// elements from hash set
return vector<int> (st.begin(), st.end());
}
int main () {
vector<int> arr = {12, 10, 9, 45, 2, 10, 10, 45};
vector<int> res = findDistinct(arr);
for (int ele: res)
cout << ele << " ";
return 0;
}
// Java program to print all distinct elements
// of a given array
import java.util.HashSet;
import java.util.ArrayList;
class GfG {
// function to return all distinct elements
static ArrayList<Integer> findDistinct(int[] arr) {
// Initialize set with all elements of array
HashSet<Integer> st = new HashSet<>();
for (int num : arr) {
st.add(num);
}
// Return the result array by inserting all
// elements from hash set
return new ArrayList<>(st);
}
public static void main(String[] args) {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};
ArrayList<Integer> res = findDistinct(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
# Python program to print all distinct elements
# of a given array
# function to return all distinct elements
def findDistinct(arr):
# Initialize set with all elements of array
st = set(arr)
# Return the result array by inserting all
# elements from hash set
return list(st)
if __name__ == "__main__":
arr = [12, 10, 9, 45, 2, 10, 10, 45]
res = findDistinct(arr)
for ele in res:
print(ele, end=" ")
// C# program to print all distinct elements
// of a given array
using System;
using System.Collections.Generic;
class GfG {
// function to return all distinct elements
static List<int> findDistinct(List<int> arr) {
// Initialize set with all elements of array
HashSet<int> st = new HashSet<int>(arr);
// Return the result array by inserting all
// elements from hash set
return new List<int>(st);
}
static void Main() {
List<int> arr = new List<int> { 12, 10, 9, 45, 2, 10, 10, 45 };
List<int> res = findDistinct(arr);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
// JavaScript program to print all distinct elements
// of a given array
// function to return all distinct elements
function findDistinct(arr) {
// Initialize set with all elements of array
const st = new Set(arr);
// Return the result array by inserting all
// elements from hash set
return Array.from(st);
}
const arr = [12, 10, 9, 45, 2, 10, 10, 45];
const res = findDistinct(arr);
console.log(res.join(" "));
Output
2 9 10 45 12
