Print sorted distinct elements of array
Last Updated :
11 Aug, 2023
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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:
// 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 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);
}
}
# 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)
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 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:
// 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 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 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# 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)
<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)
