Find all distinct elements in a given array
Last Updated :
23 Jul, 2025
Given an integer array arr[], find all the distinct elements of the given array. The given array may contain duplicates and the output should contain every element only once.
Examples:
Input: arr[] = [2, 2, 3, 3, 7, 5]
Output: [2, 3, 7, 5]
Explanation : After removing the duplicates 2 and 3 we get 2 3 7 5.
Input: arr[] = [1, 2, 3, 4, 5]
Output: [1, 2, 3, 4, 5]
Explanation : There doesn't exists any duplicate element.
[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++
#include <iostream>
#include <vector>
using namespace std;
vector<int> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
vector<int> res = remDuplicate(arr);
for (int ele : res)
cout << ele << " ";
return 0;
}
#include <stdio.h>
int* remDuplicate(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[] = {2, 2, 3, 3, 7, 5};
int size = sizeof(arr) / sizeof(arr[0]);
int resSize;
int *res = remDuplicate(arr, size, &resSize);
for (int i = 0; i < resSize; i++)
printf("%d ", res[i]);
free(res);
return 0;
}
import java.util.Arrays;
import java.util.ArrayList;
class GFG {
public ArrayList<Integer> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
ArrayList<Integer> res = remDuplicate(arr);
for (int val : res) {
System.out.print(val + " ");
}
}
}
def remDuplicate(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 = [2, 2, 3, 3, 7, 5]
res = remDuplicate(arr)
for val in res:
print(val, end=" ")
using System;
using System.Collections.Generic;
class GFG {
public List<int> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
List<int> res = remDuplicate(arr);
foreach (int val in res) {
Console.Write(val + " ");
}
}
}
function remDuplicate(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;
}
// Driver Code
let arr = [2, 2, 3, 3, 7, 5];
let res = remDuplicate(arr);
console.log(res.join(" "));
[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++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
vector<int> res = remDuplicate(arr);
for(int ele: res) {
cout << ele << " ";
}
return 0;
}
#include <stdio.h>
int compare(const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
void remDuplicate(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[] = {2, 2, 3, 3, 7, 5};
int n = sizeof(arr) / sizeof(arr[0]);
int *res;
int resSize;
remDuplicate(arr, n, &res, &resSize);
for (int i = 0; i < resSize; i++) {
printf("%d ", res[i]);
}
return 0;
}
import java.util.Arrays;
import java.util.ArrayList;
class GFG {
public ArrayList<Integer> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
ArrayList<Integer> res = remDuplicate(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
def remDuplicate(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 = [2, 2, 3, 3, 7, 5]
res = remDuplicate(arr)
for ele in res:
print(ele, end=" ")
using System;
using System.Collections.Generic;
class GFG {
public List<int> remDuplicate(int[] arr) {
// First sort the array so that all occurrences
// become consecutive
Array.Sort(arr);
var res = new List<int>();
int n = arr.Length;
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() {
int[] arr = { 2, 2, 3, 3, 7, 5 };
List<int> res = remDuplicate(arr);
foreach (int ele in res) {
Console.Write(ele + " ");
}
}
}
function remDuplicate(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;
}
// Driver Code
const arr = [2, 2, 3, 3, 7, 5];
const res = remDuplicate(arr);
console.log(res.join(" "));
[Expected Approach] Using Hashset - O(n) Time and O(n) Space
We can use Hashset 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++
#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
vector<int> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
vector<int> res = remDuplicate(arr);
for (int ele: res)
cout << ele << " ";
return 0;
}
import java.util.HashSet;
import java.util.ArrayList;
class GFG {
public ArrayList<Integer> remDuplicate(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 = {2, 2, 3, 3, 7, 5};
ArrayList<Integer> res = remDuplicate(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
def remDuplicate(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 = [2, 2, 3, 3, 7, 5]
res = remDuplicate(arr)
for ele in res:
print(ele, end=" ")
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
public List<int> remDuplicate(int[] arr) {
// Initialize HashSet with all elements of the array
var st = new HashSet<int>(arr);
// Return the result array by inserting all
// elements from hash set
return st.ToList();
}
static void Main() {
int[] arr = {2, 2, 3, 3, 7, 5};
List<int> res = remDuplicate(arr);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
function remDuplicate(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);
}
// Driver Code
const arr = [2, 2, 3, 3, 7, 5];
const res = remDuplicate(arr);
console.log(res.join(" "));
Print All Distinct Elements of a given integer array
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