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# Count distinct elements in an array

Given an unsorted array arr[] of length N, The task is to count all distinct elements in arr[].

Examples:

```Input :   arr[] = {10, 20, 20, 10, 30, 10}
Output : 3
Explanation: There are three distinct elements 10, 20, and 30.```
```Input :   arr[] = {10, 20, 20, 10, 20}
Output : 2```

Naive Approach:

Create a count variable and run two loops, one with counter i from 0 to N-1 to traverse arr[] and second with counter j from 0 to i-1 to check if ith element has appeared before. If yes, increment the count

Below is the Implementation of the above approach.

## C++

```// C++ program to count distinct elements
// in a given array
#include <iostream>
using namespace std;

int countDistinct(int arr[], int n)
{
int res = 1;

// Pick all elements one by one
for (int i = 1; i < n; i++) {
int j = 0;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;

// If not printed earlier, then print it
if (i == j)
res++;
}
return res;
}

// Driver program to test above function
int main()
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = sizeof(arr) / sizeof(arr);
cout << countDistinct(arr, n);
return 0;
}```

## Java

```// Java program to count distinct
// elements in a given array

import java.io.*;

class GFG {

static int countDistinct(int arr[], int n)
{
int res = 1;

// Pick all elements one by one
for (int i = 1; i < n; i++) {
int j = 0;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;

// If not printed earlier,
// then print it
if (i == j)
res++;
}
return res;
}

// Driver code
public static void main(String[] args)
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = arr.length;
System.out.println(countDistinct(arr, n));
}
}

// This code is contributed by Code_Mech.```

## Python3

```# Python3 program to count distinct
# elements in a given array

def countDistinct(arr, n):

res = 1

# Pick all elements one by one
for i in range(1, n):
j = 0
for j in range(i):
if (arr[i] == arr[j]):
break

# If not printed earlier, then print it
if (i == j + 1):
res += 1

return res

# Driver Code
arr = [12, 10, 9, 45, 2, 10, 10, 45]
n = len(arr)
print(countDistinct(arr, n))

# This code is contributed by Mohit Kumar
```

## C#

```// C# program to count distinct
// elements in a given array
using System;

class GFG {

static int countDistinct(int[] arr, int n)
{
int res = 1;

// Pick all elements one by one
for (int i = 1; i < n; i++) {
int j = 0;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;

// If not printed earlier,
// then print it
if (i == j)
res++;
}
return res;
}

// Driver code
public static void Main()
{
int[] arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = arr.Length;
Console.WriteLine(countDistinct(arr, n));
}
}

// This code is contributed
// by Akanksha Rai```

## Javascript

```<script>

// JavaScript program to count distinct elements
// in a given array

function countDistinct(arr, n)
{
let res = 1;

// Pick all elements one by one
for (let i = 1; i < n; i++) {
let j = 0;
for (j = 0; j < i; j++)
if (arr[i] === arr[j])
break;

// If not printed earlier, then print it
if (i === j)
res++;
}
return res;
}

// Driver program to test above function
let arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
let n = arr.length;
document.write(countDistinct(arr, n));

// This code is contributed by Surbhi Tyagi

</script>```

## PHP

```<?php
// PHP program to count distinct elements
// in a given array
function countDistinct( &\$arr, \$n)
{
\$res = 1;

// Pick all elements one by one
for ( \$i = 1; \$i < \$n; \$i++)
{

for (\$j = 0; \$j < \$i; \$j++)
if (\$arr[\$i] == \$arr[\$j])
break;

// If not printed earlier,
// then print it
if (\$i == \$j)
\$res++;
}
return \$res;
}

// Driver Code
\$arr = array( 12, 10, 9, 45, 2, 10, 10, 45 );
\$n = count(\$arr);
echo countDistinct(\$arr, \$n);

// This code is contributed by
// Rajput-Ji
?>```
Output

`5`

Time Complexity: O(n2)
Auxiliary Space: O(1)

## Count distinct elements in an array using sorting:

Below is the idea to solve the problem:

Sort the array so that all occurrences of every element become consecutive. Once the occurrences become consecutive, then traverse the sorted array and count distinct elements by comparing the consecutive elements.

Follow the steps below to Implement the idea:

• Initialize a res variable with 0 and sort arr[].
• Run a for loop from 0 to N-1.
• While i is less than N-1 and arr[i] is equal to arr[i+1] increment i.
• increment res by one.
• Return res.

Below is the implementation of above approach that is as follows:

## C++

```// C++ program to count all distinct elements
// in a given array

#include <algorithm>
#include <iostream>

using namespace std;

int countDistinct(int arr[], int n)
{
// First sort the array so that all
// occurrences become consecutive
sort(arr, arr + n);

// Traverse the sorted array
int res = 0;
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++;

res++;
}

return res;
}

// 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);
cout << countDistinct(arr, n);
return 0;
}```

## Java

```// Java program to count all distinct elements
// in a given array
import java.util.Arrays;

class GFG {

static int countDistinct(int arr[], int n)
{
// First sort the array so that all
// occurrences become consecutive
Arrays.sort(arr);

// Traverse the sorted array
int res = 0;
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++;
}
res++;
}
return res;
}

// Driver code
public static void main(String[] args)
{
int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
int n = arr.length;
System.out.println(countDistinct(arr, n));
}
}

// This code is contributed by 29AjayKumar```

## Python3

```# Python3 program to count all distinct
# elements in a given array

def countDistinct(arr, n):

# First sort the array so that all
# occurrences become consecutive
arr.sort()

# Traverse the sorted array
res = 0
i = 0
while(i < n):

# Move the index ahead while
# there are duplicates
while (i < n - 1 and
arr[i] == arr[i + 1]):
i += 1

res += 1
i += 1

return res

# Driver Code
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10]
n = len(arr)
print(countDistinct(arr, n))

# This code is contributed by mits
```

## C#

```// C# program to count all distinct elements
// in a given array
using System;

class GFG {

static int countDistinct(int[] arr, int n)
{
// First sort the array so that all
// occurrences become consecutive
Array.Sort(arr);

// Traverse the sorted array
int res = 0;
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++;
}
res++;
}
return res;
}

// Driver code
public static void Main()
{
int[] arr = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
int n = arr.Length;
Console.WriteLine(countDistinct(arr, n));
}
}

// This code is contributed by Code_Mech.```

## PHP

```<?php
// PHP program to count all distinct
// elements in a given array

function countDistinct(\$arr, \$n)
{
// First sort the array so that all
// occurrences become consecutive
sort(\$arr, 0);

// Traverse the sorted array
\$res = 0;
for (\$i = 0; \$i < \$n; \$i++)
{

// Move the index ahead while
// there are duplicates
while (\$i < \$n - 1 &&
\$arr[\$i] == \$arr[\$i + 1])
\$i++;

\$res++;
}

return \$res;
}

// Driver Code
\$arr = array( 6, 10, 5, 4, 9, 120, 4, 6, 10 );
\$n = sizeof(\$arr);
echo countDistinct(\$arr, \$n);

// This code is contributed by Akanksha Rai
?>```

## Javascript

```<script>

// JavaScript program to count all distinct elements
// in a given array

function countDistinct(arr,n)
{
// First sort the array so that all
// occurrences become consecutive
arr.sort(function(a,b){return a-b;});

// Traverse the sorted array
let res = 0;
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++;
}
res++;
}
return res;
}

// Driver code
let arr=[6, 10, 5, 4, 9, 120, 4, 6, 10];
let n = arr.length;
document.write(countDistinct(arr, n));

// This code is contributed by unknown2108

</script>```
Output

`6`

Time Complexity: O(n logn)
Auxiliary Space: O(1)

## Count distinct elements in an array usingHashing

The idea is to traverse the given array from left to right and keep track of visited elements in a hash set , as a set consists of only unique elements.

Follow the steps below to implement the idea:

1. Create an unordered_set s and a variable res initialized with 0.
2. Run a for loop from 0 to N-1
• If the current element is not present in s, insert it in s increment res by 1.
3. Return res.

Below is the implementation of the above approach.

## 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
int countDistinct(int arr[], int n)
{
// Creates an empty hashset
unordered_set<int> s;

// Traverse the input array
int res = 0;
for (int i = 0; i < n; i++) {

// If not present, then put it in
// hashtable and increment result
if (s.find(arr[i]) == s.end()) {
s.insert(arr[i]);
res++;
}
}

return res;
}

// Driver Code
int main()
{
int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
int n = sizeof(arr) / sizeof(arr);
cout << countDistinct(arr, n);
return 0;
}```

## Java

```// Java Program to count
// Unique elements in Array
import java.util.*;

class GFG {

// This method returns count
// of Unique elements
public static int countDistinct(int arr[], int n)
{

HashSet<Integer> hs = new HashSet<Integer>();

for (int i = 0; i < n; i++) {
// add all the elements to the HashSet
}

// return the size of hashset as
// it consists of all Unique elements
return hs.size();
}

// Driver code
public static void main(String[] args)
{
int arr[]
= new int[] { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
System.out.println(countDistinct(arr, arr.length));
}
}

// This code is contributed by Adarsh_Verma```

## Python3

```''' Python3 program to print all distinct elements
of a given array '''

# This function prints all distinct elements

def countDistinct(arr, n):

# Creates an empty hashset
s = set()

# Traverse the input array
res = 0
for i in range(n):

# If not present, then put it in
# hashtable and increment result
if (arr[i] not in s):
res += 1

return res

# Driver code
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10]
n = len(arr)
print(countDistinct(arr, n))

# This code is contributed by SHUBHAMSINGH10
```

## C#

```// C# Program to count
// Unique elements in Array
using System;
using System.Collections.Generic;

class GFG {

// This method returns count
// of Unique elements
public static int countDistinct(int[] arr, int n)
{

HashSet<int> hs = new HashSet<int>();

for (int i = 0; i < n; i++) {
// add all the elements to the HashSet
}

// return the size of hashset as
// it consists of all Unique elements
return hs.Count;
}

// Driver code
public static void Main()
{
int[] arr
= new int[] { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
Console.WriteLine(countDistinct(arr, arr.Length));
}
}

/* This code contributed by PrinciRaj1992 */```

## PHP

```<?php
// PHP program to print all distinct elements
// of a given array

// This function prints all distinct elements
function countDistinct(\$arr, \$n)
{
// Creates an empty hashset
\$s = array();

// Traverse the input array
\$res = 0;
for (\$i = 0; \$i < \$n; \$i++)
{
// If not present, then put it in
// hashtable and increment result
array_push(\$s,\$arr[\$i]);
}
\$s = array_unique(\$s);
return count(\$s);
}

// Driver Code
\$arr = array( 6, 10, 5, 4, 9, 120, 4, 6, 10 );
\$n = count(\$arr);
echo countDistinct(\$arr, \$n);

// This code is contributed by mits
?>```

## Javascript

```<script>
// Javascript Program to count
// Unique elements in Array

// This method returns count
// of Unique elements
function countDistinct(arr,n)
{
let hs = new Set();

for(let i = 0; i < n; i++)
{
// add all the elements to the HashSet
}

// return the size of hashset as
// it consists of all Unique elements
return hs.size;
}

// Driver code
let arr=[6, 10, 5, 4, 9,120, 4, 6, 10];
document.write(countDistinct(arr,arr.length));

// This code is contributed by patel2127
</script>```
Output

`6`

Time complexity: O(n)
Auxiliary Space: O(n), since n extra space has been taken.

## Count distinct elements in an array using Set STL:

Iterate over all the elements of the array insert them in an unordered set. As the set only contains distinct elements, so the size of set will be the answer.

Follow the below steps to Implement the idea:

• Insert all the elements into the set S one by one. • Store the total size s of the set using set::size().
• The total size s is the number of distinct elements present in the array.

Below is the Implementation of above approach.

## C++

```#include <bits/stdc++.h>
using namespace std;
// function that accepts the array and it's size and returns
// the number of distince elements
int distinct(int* arr, int len)
{

// declaring a set container using STL
set<int> S;
for (int i = 0; i < len; i++) {

// inserting all elements of the
// array into set
S.insert(arr[i]);
}
// calculating the size of the set
int ans = S.size();
return ans;
}
int main()
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
// calculating the size of the array
int l = sizeof(arr) / sizeof(int);
// calling the function on array
int dis_elements = distinct(arr, l);
cout << dis_elements << endl;
return 0;
}```

## Java

```import java.util.*;
class GFG {

// function that accepts
// the array and it's size and
// returns the number of distince elements
static int distinct(int[] arr, int len)
{
// declaring a set container
// using STL
HashSet<Integer> S = new HashSet<>();

for (int i = 0; i < len; i++) {
// inserting all elements of the
// array into set
}

// calculating the size of the set
int ans = S.size();
return ans;
}

// Driver code
public static void main(String[] args)
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };

// calculating the size of the
// array
int l = arr.length;
// calling the function on array
int dis_elements = distinct(arr, l);
System.out.print(dis_elements + "\n");
}
}

// This code is contributed by Rajput-Ji```

## Python3

```# function that accepts the array and it's size and returns
# the number of distince elements

def distinct(arr, l):
# declaring a set container using STL
S = set()
for i in range(l):
# inserting all elements of the
# array into set

# calculating the size of the set
ans = len(S)
return ans

# Driver code
if __name__ == '__main__':
arr = [12, 10, 9, 45, 2, 10, 10, 45]

# calculating the size of the array
l = len(arr)

# calling the function on array
dis_elements = distinct(arr, l)
print(dis_elements, "")

# This code is contributed by Rajput-Ji
```

## C#

```using System;
using System.Collections.Generic;

public class GFG {

// function that accepts the array and it's size and
// returns the number of distince elements
static int distinct(int[] arr, int len)
{
// declaring a set
// container using STL
HashSet<int> S = new HashSet<int>();
for (int i = 0; i < len; i++) {
// inserting all elements of the
// array into set
}
// calculating the size of the set
int ans = S.Count;
return ans;
}

// Driver code
public static void Main(String[] args)
{
int[] arr = { 12, 10, 9, 45, 2, 10, 10, 45 };

// calculating the size of the array
int l = arr.Length;

// calling the function on array
int dis_elements = distinct(arr, l);
Console.Write(dis_elements + "\n");
}
}

// This code is contributed by Rajput-Ji```

## Javascript

```<script>
// function that accepts the array and it's size and returns
// the number of distince elements
function distinct(arr , len)
{
var S = new Set(); // declaring a set container using STL
var i =0;
for (i = 0; i < len; i++)
{
S.add(arr[i]); // inserting all elements of the
// array into set
}
var ans = S.size; // calculating the size of the set
return ans;
}

// Driver code
var arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
var l = arr.length; // calculating the size of the array
var dis_elements = distinct(arr, l); // calling the function on array
document.write(dis_elements);

// This code is contributed by Rajput-Ji.
</script>```
Output

`5`

## Python3

```arr = [12, 10, 9, 45, 2, 10, 10, 45]
print(len(set(arr)))
```

## Javascript

```let arr = [12, 10, 9, 45, 2, 10, 10, 45]
console.log((new Set(arr)).size)```

## Java

```import java.util.HashSet;

public class Program {

// Driver Code
public static void main(String[] args)
{
int[] arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
HashSet<Integer> set = new HashSet<>();
for (int i = 0; i < arr.length; i++) {
}
System.out.println(set.size());
}
}```
Output

```5
```

### Count distinct elements in an array using HASHMAP STL:

Create the map and store the elements in the map with value as their frequency because duplicate cannot exist in map data structure. So all the values that has been inserted into the map will be distinct. Finally. the size of the map will give you the number of distinct elements in the array present in the given input array(or vector).

Implementation of the above Approach is given below:

## C++

```#include <bits/stdc++.h>
using namespace std;

int main() {
vector<int> arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
map<int, int> distinct;
// here we are storing all the elements into
// the map data structure
for(int i = 0; i < arr.size(); i++){
distinct[arr[i]]++;
}
cout << distinct.size();
}

// This code is contributed by Prince Kumar```

## Java

```import java.util.*;

public class Main {
public static void main(String[] args) {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};

Map<Integer, Integer> distinct = new HashMap<>();

// Here we are storing all the elements into
// the map data structure
for (int num : arr) {
distinct.put(num, distinct.getOrDefault(num, 0) + 1);
}

System.out.println(distinct.size());
}
}
```

## Python

```# Python code to find the number
# of distinct elements in an array
# using Dictionary and without
# using any in built module

# defining the array
arr = [12,10,9,45,2,10,10,45]

# Initializing the blank dictionary
dic = {}

# Iterating over each element of the array arr
for i in arr:

# Checking if the element is present in
# the dictionary or not

# If not then add it to the dictionary
# and make the value 1
if i not in dic:
dic[i] = 1
# If present already then just increase the
# value by 1
else:
dic[i] += 1

# Printing the length of the keys of the dictionary
print(len(dic.keys()))```

## C#

```using System;
using System.Collections.Generic;

class MainClass {
public static void Main (string[] args) {
int[] arr = {12, 10, 9, 45, 2, 10, 10, 45};

Dictionary<int, int> distinct = new Dictionary<int, int>();

// Here we are storing all the elements into
// the map data structure
foreach (int num in arr) {
if (distinct.ContainsKey(num)) {
distinct[num]++;
} else {
distinct[num] = 1;
}
}

Console.WriteLine(distinct.Count);
}
}```

## Javascript

```let arr = [12, 10, 9, 45, 2, 10, 10, 45];
let distinct = new Map();

// Here we are storing all the elements into
// the map data structure
for (let i = 0; i < arr.length; i++) {
let current = arr[i];
if (distinct.has(current)) {
distinct.set(current, distinct.get(current) + 1);
} else {
distinct.set(current, 1);
}
}

console.log(distinct.size);
```

## Python3

```# Python program for the above approach

from collections import defaultdict

arr = [12, 10, 9, 45, 2, 10, 10, 45]

distinct = defaultdict(int)

# here we are storing all the elements into
# the map data structure
for num in arr:
distinct[num] += 1

print(len(distinct))

# This code is contributed by rishabmalhdjio```
Output

`5`

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

My Personal Notes arrow_drop_up