Find all distinct elements in a given array

Last Updated : 07 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.

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++

#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; }

Java

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 + " ");  }  } }

Python

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 + " ");  }  } }

JavaScript

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(" "));

Output

2 3 7 5

[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; }

Java

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 + " ");  }  } }

Python

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 + " ");  }  } }

JavaScript

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(" "));

Output

2 3 5 7

[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; }

Java

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 + " ");  }  } }

Python

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 + " ");  } }

JavaScript

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(" "));