Print sorted distinct elements of array (original) (raw)

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:

  1. Sort the given array in non-descending order.
  2. 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.
  3. 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);

`

**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 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

`

**Complexity Analysis:

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