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)