# Java program to count distinct element in an array

## Count distinct element in an array in Java

In this section, we will learn, how to Count distinct element in an array in java language.

Given an integer array, we have to print all the distinct element of the input array. input array may contain duplicate elements, we have to print the count of distinct elements.

## Methods Discussed are :

• Method 1 : Using Two loops
• Method 2 : Using hash Map

Let’s discuss each method one by one,

## Method 1 :

In this method we will count the frequency of each elements using two for loops.

• To check the status of visited elements create a array of size n and a variable say count_dis=0 to count the distinct element.
• Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
• Run a loop from index i+1 to n
• Check if(arr[i]==arr[j]), then set visited[j]=1.
• After complete iteration of inner for loop and increment the value of count_dis
• At last print the value count_dis

### Time and Space Complexity :

• Time Complexity : O(n2)
• Space Complexity : O(n)

### Method 1 : Code in Java

Run
```import java.util.Arrays;

class Main
{
public static void countFreq(int arr[], int n)
{
boolean visited[] = new boolean[n];
Arrays.fill(visited, false);
int count_dis=0;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++) {

// Skip this element if already processed
if (visited[i] == true)
continue;

for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
visited[j] = true;

}
}
count_dis = count_dis+1;
}
System.out.println(count_dis);
}

// Driver code
public static void main(String []args)
{
int arr[] = new int[]{10, 30, 40, 20, 10, 20, 50, 10};
int n = arr.length;
countFreq(arr, n);
}
}```

`5`

## Method 2 :

In this method we will count use hash-map to count the frequency of each elements.

• Declare a hash map and a variable count_dis=0.
• Start iterating over the entire array
• If element is present in map, then increase the value of frequency by 1.
• Otherwise, insert that element in map.
• After complete iteration over array, print the size of map.

### Time and Space Complexity :

• Time Complexity : O(n)
• Space Complexity : O(n)

### Method 2 : Code in Java

Run
```import java.util.*;
import java.util.Arrays;

class Main
{
static void countFreq(int arr[], int n)
{
Map<Integer, Integer> mp = new HashMap<>();
int count_dis=0;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
{
if (mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i]) + 1);
}
else
{
mp.put(arr[i], 1);
}
}

System.out.println(mp.size());
}
// Driver code
public static void main(String []args)
{
int arr[] = new int[]{10, 40, 50, 20, 10, 20, 30, 10};
int n = arr.length;
countFreq(arr, n);
}
}```

`5`

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### 9 comments on “Java program to count distinct element in an array”

• Faiz

import java.util.*;

public class bari {

public static void solution(int[] arr) {
HashSet set = new HashSet();

for (int i : arr) {
}

int count = set.size();

System.out.println(“Distinct element in array is : ” + count);
}

public static void main(String[] args) {
int arr[] = new int[] { 10, 40, 50, 20, 10, 20, 30, 10 };

solution(arr);
}
}

• Ritik

class Main{
static int distinctElements(int[] ia){
int count=1;
java.util.Arrays.sort(ia);

for (int i=0,j=1;i<ia.length && j<ia.length ;i++ , j++)
{
if(ia[i]==ia[j]){
continue;
}
count++;
}
return count;
}

public static void main(String[] args){
int[] ia={4,5,3,7,98,0,2,2,3,5,5,6};
System.out.println("number of Distinct elements :"+distinctElements(ib));
}
}

• Debyanshu

import java.util.*;

public class Main {
public static void main(String[] args) {
int a[]={10, 30, 40, 20, 10, 20, 50, 10};
System.out.println(map(a));

}
static int map(int a[]) {

HashSet map=new HashSet();
for (int i = 0; i <a.length ; i++) {

}
return map.size();

}
}

• Ganpat

// Finding Distinct numbers from the array
//Easy method
import java.util.*;
class Main{
public static void main(String [] args){

int arr[] = {10,20,50,40,50,45,10,60,20};
int len = arr.length;
Arrays.sort(arr);

int count=1;
for(int i=0; i<len-1; i++){
if(arr[i] == arr[i+1]){
continue;
}
else{
count++;
}
}
System.out.println(count);
}
}

• Gyanendra

import java.util.*;
public class Find_distinct_element_in_an_array {

public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++) {
arr[i]=scn.nextInt();
}

Map list=new HashMap();

for(int i=0;i<arr.length;i++) {
list.put(arr[i],i);
}
System.out.println(list.size());

}

}

• 19cs134-

import java.util.*;

public class Main {
public static void main(String[] args) {
HashSet h= new HashSet();
Scanner s = new Scanner(System.in);
int q=5;
for(int i=0;i< a;i++){
int temp = s.nextInt();
}
System.out.println(h.size());
}
}

• Dhivya

import java.util.HashSet;
import java.util.Scanner;

public class distinctArray {
public static void main(String[] args) {
HashSet h= new HashSet();
Scanner s = new Scanner(System.in);
int a[]= new int[5];
for(int i=0;i< a.length;i++){
a[i]= s.nextInt();

}
for(int i=0;i< a.length;i++){