# Print All Distinct Elements of a given integer array

Given an integer array, print all distinct elements in array. The given array may contain duplicates and the output should print every element only once. The given array is not sorted.

Examples:

```Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 12, 10, 9, 45, 2

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1, 2, 3, 4, 5

Input: arr[] = {1, 1, 1, 1, 1}
Output: 1```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Simple Solution is to use twp 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 ignores the element, else prints the element. Following is C++ implementation of the simple algorithm.

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

void printDistinct(int arr[], int n)
{
// Pick all elements one by one
for (int i=0; i<n; i++)
{
// Check if the picked element is already printed
int j;
for (j=0; j<i; j++)
if (arr[i] == arr[j])
break;

// If not printed earlier, then print it
if (i == j)
cout << arr[i] << " ";
}
}

// 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[0]);
printDistinct(arr, n);
return 0;
}```

Output:

`6 10 5 4 9 120`

Time Complexity of above solution is O(n2). We can Use Sorting to solve the problem in O(nLogn) time. The idea is simple, first 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 in O(n) time. Following is C++ implementation of the idea.

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

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

// Traverse the sorted array
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++;

// print last occurrence of the current element
cout << arr[i] << " ";
}
}

// 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[0]);
printDistinct(arr, n);
return 0;
}
```

Output:

`4 5 6 9 10 120`

We can Use Hashing to solve this in O(n) time on average. The idea is to traverse the given array from left to right and keep track of visited elements in a hash table. Following is Java implementation of the idea.

```/* Java program to print all distinct elements of a given array */
import java.util.*;

class Main
{
// This function prints all distinct elements
static void printDistinct(int arr[])
{
// Creates an empty hashset
HashSet<Integer> set = new HashSet<>();

// Traverse the input array
for (int i=0; i<arr.length; i++)
{
// If not present, then put it in hashtable and print it
if (!set.contains(arr[i]))
{
set.add(arr[i]);
System.out.print(arr[i] + " ");
}
}
}

// Driver method to test above method
public static void main (String[] args)
{
int arr[] = {10, 5, 3, 4, 3, 5, 6};
printDistinct(arr);
}
}
```

Output:

`10 5 3 4 6 `

One more advantage of hashing over sorting is, the elements are printed in same order as they are in input array.

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