# Java program to find non-repeating elements in an array

## Non Repeating elements in an Array in Java

Here, in this page you will find the code for printing non repeating elements in an array in java programming language We are given with an array and need to print the distinct elements among them. You will found different methods to solve this problem in this page.

Here, we will discuss the following methods in this page to print the distinct elements in the given input array.

• 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.
• Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
• Otherwise create a variable count = 1 to keep the count of frequency.
• Run a loop from index i+1 to n
• Check if(arr[i]==arr[j]), then increment the count by 1 and set visited[j]=1.
• After complete iteration of inner for loop and check if count==1, then print that ith element.

### 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);
// 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;

// Count frequency
int count = 1;
for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
visited[j] = true;
count++;
}
}
if(count==1)
System.out.println(arr[i]);
}
} // 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);
}
}
```

### Output

`30 40 50`

## 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, start traversing map and print those key which have value 1.

### 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);
}
}
// Traverse through map and print frequencies
for (Map.Entry<Integer, Integer> entry : mp.entrySet())
{    if(entry.getValue()==1)
System.out.println(entry.getKey()+" ");
}

}
// 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);
}
}```

### Output

`30 40 50`

### One comment on “Java program to find non-repeating elements in an array”

• Mahesh

int[]a={1,2,42,13,3,55,3,23,55,45,6,25,5,5,53,34,34,33,35};
Arrays.sort(a);
boolean []vis=new boolean[a.length];
for(int i=0;i<a.length-1;i++){
if(a[i]==a[i+1]){
vis[i]=true;
vis[i+1]=true;
}
}
for(int i=0;i<a.length;i++){
if(!vis[i]){
System.out.print(a[i]+" ");
}
}