Print sorted distinct elements of array

Last Updated : 11 Aug, 2023

Given an array that might contain duplicates, print all distinct elements in sorted order.

Examples:

Input  : 1, 3, 2, 2, 1
Output : 1 2 3
Input  : 1, 1, 1, 2, 2, 3
Output : 1 2 3

The simple Solution is to sort the array first, then traverse the array and print only first occurrences of elements.

Algorithm:

Sort the given array in non-descending order.
Traverse the sorted array from left to right, and for each element do the following:
a. If the current element is not equal to the previous element, print the current element.
b. Otherwise, skip the current element and move on to the next one.
Stop when the end of the array is reached.

Below is the implementation of the approach:

C++

   // CPP program to print sorted distinct
// elements.
 
#include <bits/stdc++.h> 
using namespace std; 
 
// Function to print sorted distinct
// elements
void printDistinct(int arr[], int n) { 
    // Sort the array 
    sort(arr, arr + n); 
 
    // Traverse the sorted array 
    for (int i = 0; i < n; i++) { 
 
        // If the current element is not equal to the previous 
        // element, print it 
        if (i == 0 || arr[i] != arr[i - 1]) 
            cout << arr[i] << " "; 
    } 
} 
 
// Driver's code
int main() { 
      // Input array
    int arr[] = { 1, 3, 2, 2, 1 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
   
      // Function call
    printDistinct(arr, n); 
   
    return 0; 
} 
 
  

Java

   // Java program to print sorted distinct
// elements.
 
import java.util.*;
 
class GFG {
    // Function to print sorted distinct
    // elements
    static void printDistinct(int arr[], int n) {
        // Sort the array
        Arrays.sort(arr);
       
        // Traverse the sorted array
        for (int i = 0; i < n; i++) {
            // If the current element is not equal to the
            // previous element, print it
            if (i == 0 || arr[i] != arr[i - 1])
                System.out.print(arr[i] + " ");
        }
    }
 
    // Driver's code
    public static void main(String[] args) {
        // Input array
        int arr[] = { 1, 3, 2, 2, 1 };
        int n = arr.length;
 
        // Function call
        printDistinct(arr, n);
    }
}
 
  

Python3

   # Python program to print sorted distinct elements
def printDistinct(arr):
    # Sort the array
    arr.sort()
 
    # Traverse the sorted array
    for i in range(len(arr)):
 
        # If the current element is not equal to the previous
        # element, print it
        if i == 0 or arr[i] != arr[i - 1]:
            print(arr[i])
 
# Driver code
arr = [1, 3, 2, 2, 1]
printDistinct(arr)
 
  

C#

   using System;
using System.Linq;
 
class GFG {
    // Function to print sorted distinct elements
    static void printDistinct(int[] arr, int n) {
        // Sort the array
        Array.Sort(arr);
 
        // Traverse the sorted array
        for (int i = 0; i < n; i++) {
            // If the current element is not equal to the previous element, print it
            if (i == 0 || arr[i] != arr[i - 1])
                Console.Write(arr[i] + " ");
        }
    }
 
    // Driver's code
    public static void Main() {
        // Input array
        int[] arr = { 1, 3, 2, 2, 1 };
        int n = arr.Length;
 
        // Function call
        printDistinct(arr, n);
    }
}
// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL
 
  

Javascript

   // Javascript program to print sorted distinct
// elements.
function printDistinct(arr) {
    // Sort the array
    arr.sort((a, b) => a - b);
 
    // Traverse the sorted array
    for (let i = 0; i < arr.length; i++) {
 
        // If the current element is not equal to the previous
        // element, print it
        if (i === 0 || arr[i] !== arr[i - 1]) {
            console.log(arr[i]);
        }
    }
}
 
// Driver code
const arr = [1, 3, 2, 2, 1];
printDistinct(arr);
 
  

Output

1 2 3

Time Complexity: O(n * logn) as sort function has been called which takes O(n * logn) time. Here, n is size of the input array.

Auxiliary Space: O(1) as no extra space has been used.

Another Approach is to use set in C++ STL.

Implementation:

C++

   // CPP program to print sorted distinct
// elements.
#include <bits/stdc++.h>
using namespace std;
 
void printRepeating(int arr[], int size)
{
    // Create a set using array elements
    set<int> s(arr, arr + size);
 
    // Print contents of the set.
    for (auto x : s) 
        cout << x << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printRepeating(arr, n);
    return 0;
}
 
  

Java

   // Java program to print sorted distinct
// elements.
import java.io.*;
import java.util.*;
 
public class GFG {
      
    static void printRepeating(Integer []arr, int size)
    {
        // Create a set using array elements
        SortedSet<Integer> s = new TreeSet<>();
        Collections.addAll(s, arr);
         
        // Print contents of the set.
        System.out.print(s);
    }
      
    // Driver code
    public static void main(String args[])
    {
        Integer []arr = {1, 3, 2, 2, 1};
        int n = arr.length;
        printRepeating(arr, n);
    }
}
  
// This code is contributed by
// Manish Shaw (manishshaw1)
 
  

Python3

   # Python3 program to print 
# sorted distinct elements.
 
def printRepeating(arr,size):
 
    # Create a set using array elements 
    s = set()
    for i in range(size):
        if arr[i] not in s:
            s.add(arr[i])
 
    # Print contents of the set.
    for i in s:
        print(i,end=" ")
 
# Driver code
if __name__=='__main__':
    arr = [1,3,2,2,1]
    size = len(arr)
    printRepeating(arr,size)
 
# This code is contributed by 
# Shrikant13
 
  

C#

   // C# program to print sorted distinct
// elements.
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG {
     
    static void printRepeating(int []arr, int size)
    {
        // Create a set using array elements
        SortedSet<int> s = new SortedSet<int>(arr);
     
        // Print contents of the set.
        foreach (var n in s)
        {
            Console.Write(n + " ");
        } 
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = {1, 3, 2, 2, 1};
        int n = arr.Length;
        printRepeating(arr, n);
    }
}
 
// This code is contributed by
// Manish Shaw (manishshaw1)
 
  

Javascript

   <script>
 
// Javascript program to print sorted distinct
// elements.
function printRepeating(arr, size)
{
     
    // Create a set using array elements
    var s = new Set(arr);
 
    // Print contents of the set.
    [...s].sort((a, b) => a - b).forEach(x => {
        document.write(x + " ")
    });
}
 
// Driver code
var arr = [ 1, 3, 2, 2, 1 ];
var n = arr.length;
 
printRepeating(arr, n);
 
// This code is contributed by itsok
 
</script>
 
  

Output

1 2 3

Complexity Analysis:

Time Complexity: O(nlogn).
Auxiliary Space: O(n)

Print all Distinct (Unique) Elements in given Array

Anivesh Tiwari

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Print sorted distinct elements of array

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