Elements that occurred only once in the array

Given an array arr that has numbers appearing twice or once. The task is to identify numbers that occur only once in the array. 

Note: Duplicates appear side by side every time. There might be a few numbers that can occur at one time and just assume this is a right rotating array (just say an array can rotate k times towards right). The order of the elements in the output doesn’t matter.

Examples:

Input: arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 }
Output: 9 4

Input: arr[] = {-9, -8, 4, 4, 5, 5, -1}
Output: -9 -8 -1

Method-1: Using Sorting.

• Sort the array.
• Check for each element at index i (except the first and last element), if

arr[i] != arr[i-1] && arr [i] != arr[i+1]

• For the first element, check if arr[0] != arr[1].
• For the last element, check if arr[n-1] != arr[n-2].

Algorithm:

1. Sort the given array in non-decreasing order using any sorting algorithm.
2. Traverse the sorted array and compare each element with its adjacent element.
3. If an element is not equal to its adjacent elements, then print it.
4. For the first element, check if it is different from the second element. If yes, print it.
5. For the last element, check if it is different from the second last element. If yes, print it.
   

Pseudocode:

occurredOnce(arr[], n):
sort(arr, arr + n)
for i = 0 to n-1 do
if i == 0 and arr[i] != arr[i+1] then
print arr[i]
else if i == n-1 and arr[i] != arr[i-1] then
print arr[i]
else if arr[i] != arr[i-1] and arr[i] != arr[i+1] then
print arr[i]

Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);

    // Check for first element
    if (arr[0] != arr[1])
        cout << arr[0] << " ";

    // Check for all the elements if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] && arr[i] != arr[i - 1])
            cout << arr[i] << " ";

    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        cout << arr[n - 1] << " ";
}

// Driver code
int main()
{

    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    occurredOnce(arr, n);

    return 0;
}

Java

// Java implementation
// of above approach
import java.util.*;

class GFG
{

// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int arr[], int n)
{
    // Sort the array
    Arrays.sort(arr);

    // Check for first element
    if (arr[0] != arr[1])
        System.out.println(arr[0] + " ");

    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            System.out.print(arr[i] + " ");

    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        System.out.print(arr[n - 1] + " ");
}

// Driver code
public static void main(String args[])
{
    int arr[] = {7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2};
    int n = arr.length;

    occurredOnce(arr, n);
}
}

// This code is contributed
// by Arnab Kundu

Python3

# Python 3 implementation
# of above approach

# Function to find the elements
# that appeared only once in
# the array
def occurredOnce(arr, n):

    # Sort the array
    arr.sort()

    # Check for first element
    if arr[0] != arr[1]:
        print(arr[0], end = " ")

    # Check for all the elements
    # if it is different its
    # adjacent elements
    for i in range(1, n - 1):
        if (arr[i] != arr[i + 1] and
            arr[i] != arr[i - 1]):
            print( arr[i], end = " ")

    # Check for the last element
    if arr[n - 2] != arr[n - 1]:
        print(arr[n - 1], end = " ")

# Driver code
if __name__ == "__main__":

    arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ]
    n = len(arr)
    occurredOnce(arr, n)

# This code is contributed
# by ChitraNayal

Javascript

<script>

// Javascript implementation
// of above approach

// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr,n)
{
    // Sort the array
    arr.sort(function(a,b){return a-b;});

    // Check for first element
    if (arr[0] != arr[1])
        document.write(arr[0] + " ");

    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (let i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            document.write(arr[i] + " ");

    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        document.write(arr[n - 1] + " ");
}

// Driver code
let arr=[7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2];
let n = arr.length;
occurredOnce(arr, n);


// This code is contributed by rag2127

</script>

C#

// C# implementation
// of above approach
using System;

class GFG
{

// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int[] arr, int n)
{
    // Sort the array
    Array.Sort(arr);

    // Check for first element
    if (arr[0] != arr[1])
        Console.Write(arr[0] + " ");

    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            Console.Write(arr[i] + " ");

    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        Console.Write(arr[n - 1] + " ");
}

// Driver code
public static void Main()
{
    int[] arr = {7, 7, 8, 8, 9,
                1, 1, 4, 2, 2};
    int n = arr.Length;

    occurredOnce(arr, n);
}
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP implementation
// of above approach

// Function to find the elements
// that appeared only once in
// the array
function occurredOnce(&$arr, $n)
{
    // Sort the array
    sort($arr);

    // Check for first element
    if ($arr[0] != $arr[1])
        echo $arr[0]." ";

    // Check for all the elements
    // if it is different its
    // adjacent elements
    for ($i = 1; $i < $n - 1; $i++)
        if ($arr[$i] != $arr[$i + 1] &&
            $arr[$i] != $arr[$i - 1])
            echo $arr[$i]." ";

    // Check for the last element
    if ($arr[$n - 2] != $arr[$n - 1])
        echo $arr[$n - 1]." ";
}

// Driver code
$arr = array(7, 7, 8, 8, 9,
             1, 1, 4, 2, 2);
$n = sizeof($arr);
occurredOnce($arr, $n);

// This code is contributed
// by ChitraNayal
?>

Output

4 9 

Complexity Analysis:

• Time Complexity: O(Nlogn) 
• Auxiliary Space: O(1)

Method-2: (Using Hashing): In C++, unordered_map can be used for hashing. 

• Traverse the array.
• Store each element with its occurrence in the unordered_map.
• Traverse the unordered_map and print all the elements with occurrence 1.

Below is the implementation of the above approach: 

C++

// C++ implementation to find elements
// that appeared only once
#include <bits/stdc++.h>
using namespace std;

// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    unordered_map<int, int> mp;

    // Store all the elements in the map with
    // their occurrence
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;

    // Traverse the map and print all the
    // elements with occurrence 1
    for (auto it = mp.begin(); it != mp.end(); it++)
        if (it->second == 1)
            cout << it->first << " ";
}

// Driver code
int main()
{

    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    occurredOnce(arr, n);

    return 0;
}

Java

// Java implementation to find elements
// that appeared only once
import java.util.*;
import java.io.*;

class GFG
{
    // Function to find the elements that
    // appeared only once in the array
    static void occurredOnce(int[] arr, int n)
    {
            HashMap<Integer, Integer> mp = new HashMap<>();

            // Store all the elements in the map with
            // their occurrence
            for (int i = 0; i < n; i++)
            {
                if (mp.containsKey(arr[i]))
                    mp.put(arr[i], 1 + mp.get(arr[i]));
                else
                    mp.put(arr[i], 1);
            }

            // Traverse the map and print all the
            // elements with occurrence 1
            for (Map.Entry entry : mp.entrySet())
            {
                if (Integer.parseInt(String.valueOf(entry.getValue())) == 1)
                    System.out.print(entry.getKey() + " ");
            }
    }

    // Driver code
    public static void main(String args[])
    {
            int[] arr = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
            int n = arr.length;

            occurredOnce(arr, n);
    }
}

// This code is contributed by rachana soma

Python3

# Python3 implementation to find elements
# that appeared only once
import math as mt

# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):

    mp = dict()

    # Store all the elements in the
    # map with their occurrence
    for i in range(n):
        if arr[i] in mp.keys():
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1

    # Traverse the map and print all
    # the elements with occurrence 1
    for it in mp:
        if mp[it] == 1:
            print(it, end = " ")

# Driver code
arr = [7, 7, 8, 8, 9, 1, 1, 4, 2, 2]
n = len(arr)

occurredOnce(arr, n)

# This code is contributed by
# Mohit Kumar 29

Javascript

<script>

// Javascript implementation to find elements
// that appeared only once

// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr, n)
{
    let mp = new Map();

    // Store all the elements in the map
    // with their occurrence
    for(let i = 0; i < n; i++)
    {
        if (mp.has(arr[i]))
            mp.set(arr[i], 1 + mp.get(arr[i]));
        else
            mp.set(arr[i], 1);
    }

    // Traverse the map and print all the
    // elements with occurrence 1
    for(let [key, value] of mp.entries())
    {
        if (value == 1)
            document.write(key + " ");
    }
}

// Driver code
let arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ];
let n = arr.length;

occurredOnce(arr, n);

// This code is contributed by avanitrachhadiya2155

</script>

C#

// C# implementation to find elements
// that appeared only once
using System;
using System.Collections.Generic;
class GFG
{
    // Function to find the elements that
    // appeared only once in the array
    static void occurredOnce(int[] arr, int n)
    {
      Dictionary<int, int> mp = new Dictionary<int, int>();

      // Store all the elements in the map with
      // their occurrence
      for (int i = 0; i < n; i++)
      {
        if (mp.ContainsKey(arr[i]))
          mp[arr[i]] = 1 + mp[arr[i]];
        else
          mp.Add(arr[i], 1);
      }

      // Traverse the map and print all the
      // elements with occurrence 1
      foreach(KeyValuePair<int, int> entry in mp)
      {
        if (Int32.Parse(String.Join("", entry.Value)) == 1)
          Console.Write(entry.Key + " ");
      }
    }

    // Driver code
    public static void Main(String []args)
    {
      int[] arr = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
      int n = arr.Length;
      occurredOnce(arr, n);
    }
}

// This code is contributed by shikhasingrajput

Output

4 9 

Complexity Analysis:

• Time Complexity: O(N) 
• Auxiliary Space: O(N)

Method-3: Using given assumptions.

It is given that an array can be rotated any time and duplicates will appear side by side every time. So, after rotating, the first and last elements will appear side by side. 

• Check if the first and last elements are equal. If yes, then start traversing the elements between them.
• Check if the current element is equal to the element in the immediate previous index. If yes, check the same for the next element.
• If not, print the current element.

Implementation:

C++

// C++ implementation to find elements
// that appeared only once
#include <bits/stdc++.h>
using namespace std;

// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    int i = 1, len = n;

    // Check if the first and last element is equal
    // If yes, remove those elements
    if (arr[0] == arr[len - 1]) {
        i = 2;
        len--;
    }

    // Start traversing the remaining elements
    for (; i < n; i++)

        // Check if current element is equal to
        // the element at immediate previous index
        // If yes, check the same for next element
        if (arr[i] == arr[i - 1])
            i++;

        // Else print the current element
        else
            cout << arr[i - 1] << " ";

    // Check for the last element
    if (arr[n - 1] != arr[0] && arr[n - 1] != arr[n - 2])
        cout << arr[n - 1];
}

// Driver code
int main()
{

    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    occurredOnce(arr, n);

    return 0;
}

Java

// Java implementation to find
// elements that appeared only once
class GFG
{
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int arr[], int n)
{
    int i = 1, len = n;

    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }

    // Start traversing the
    // remaining elements
    for (; i < n; i++)

        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;

        // Else print the current element
        else
            System.out.print(arr[i - 1] + " ");

    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        System.out.print(arr[n - 1]);
}

// Driver code
public static void main(String args[])
{
    int arr[] = {7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2};
    int n = arr.length;

    occurredOnce(arr, n);
}
}

// This code is contributed
// by Arnab Kundu

Python3

# Python 3 implementation to find
# elements that appeared only once

# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):
    i = 1
    len = n

    # Check if the first and
    # last element is equal
    # If yes, remove those elements
    if arr[0] == arr[len - 1]:
        i = 2
        len -= 1

    # Start traversing the
    # remaining elements
    while i < n:

        # Check if current element is
        # equal to the element at
        # immediate previous index
        # If yes, check the same for
        # next element
        if arr[i] == arr[i - 1]:
            i += 1

        # Else print the current element
        else:
            print(arr[i - 1], end = " ")

        i += 1

    # Check for the last element
    if (arr[n - 1] != arr[0] and
        arr[n - 1] != arr[n - 2]):
        print(arr[n - 1])

# Driver code
if __name__ == "__main__":
    arr = [ 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 ]
    n = len(arr)

    occurredOnce(arr, n)

# This code is contributed
# by ChitraNayal

Javascript

<script>

// Javascript implementation to find
// elements that appeared only once

// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr, n)
{
    var i = 1, len = n;

    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }

    // Start traversing the
    // remaining elements
    for(; i < n; i++)

        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;

        // Else print the current element
        else
            document.write(arr[i - 1] + " ");

    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        document.write(arr[n - 1]);
}

// Driver code
var arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ];
var n = arr.length;

occurredOnce(arr, n);

// This code is contributed by Ankita saini

</script>

C#

// C# implementation to find
// elements that appeared only once
using System;

class GFG
{
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int[] arr, int n)
{
    int i = 1, len = n;

    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }

    // Start traversing the
    // remaining elements
    for (; i < n; i++)

        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;

        // Else print the current element
        else
            Console.Write(arr[i - 1] + " ");

    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        Console.Write(arr[n - 1]);
}

// Driver code
public static void Main()
{
    int[] arr = {7, 7, 8, 8, 9,
                1, 1, 4, 2, 2};
    int n = arr.Length;

    occurredOnce(arr, n);
}
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP implementation to find
// elements that appeared only once

// Function to find the elements that
// appeared only once in the array
function occurredOnce(&$arr, $n)
{
    $i = 1;
    $len = $n;

    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if ($arr[0] == $arr[$len - 1])
    {
        $i = 2;
        $len--;
    }

    // Start traversing the
    // remaining elements
    for (; $i < $n; $i++)

        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same for
        // next element
        if ($arr[$i] == $arr[$i - 1])
            $i++;

        // Else print the current element
        else
            echo $arr[$i - 1] . " ";

    // Check for the last element
    if ($arr[$n - 1] != $arr[0] &&
        $arr[$n - 1] != $arr[$n - 2])
        echo $arr[$n - 1];
}

// Driver code
$arr = array(7, 7, 8, 8, 9,
             1, 1, 4, 2, 2);
$n = sizeof($arr);

occurredOnce($arr, $n);

// This code is contributed
// by ChitraNayal
?>

Output

9 4 

Complexity Analysis:

• Time Complexity: O(N) 
• Auxiliary Space: O(1)

Method #4:Using built-in Python functions:

• Count the frequencies of every element using the Counter function
• Traverse the frequency array and print all the elements with occurrence 1.

Below is the implementation

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find the elements that
// appeared only once in the array
void OccurredOnce(int arr[], int n) {
    // counting frequency of every element
    // using unordered_map
    unordered_map<int, int> mp;
    for(int i=0;i<n;i++){
        mp[arr[i]]++;
    }

    // Traverse the map and print all
    // the elements with occurrence 1
    for(auto item: mp){
        if (item.second == 1) {
            cout<<item.first<<" ";
        }
    }
}

// Driver code
int main() {
    int arr[] = {-9, -8, 4, 4, 5, 5, -1};
    int n = sizeof(arr)/sizeof(arr[0]);
    OccurredOnce(arr, n);
    return 0;
}

// This code is contributed by adityashatmfh

Java

import java.util.HashMap;

public class Main {
    public static void OccurredOnce(int[] arr) {
        // Counting frequency of every element using HashMap
        HashMap<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            if (mp.containsKey(arr[i])) {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            } else {
                mp.put(arr[i], 1);
            }
        }
        // Traverse the map and print all the elements with occurrence 1
        for (int it : mp.keySet()) {
            if (mp.get(it) == 1) {
                System.out.print(it + " ");
            }
        }
    }

    public static void main(String[] args) {
        int[] arr = {7, 7, 8, 8, 9, 1, 1, 4, 2, 2};
        OccurredOnce(arr);
    }
}

Python3

# Python3 implementation to find elements
# that appeared only once
from collections import Counter

# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):

    #counting frequency of every element using Counter
    mp=Counter(arr)
    # Traverse the map and print all
    # the elements with occurrence 1
    for it in mp:
        if mp[it] == 1:
            print(it, end = " ")

# Driver code
arr = [7, 7, 8, 8, 9, 1, 1, 4, 2, 2]
n = len(arr)

occurredOnce(arr, n)

# This code is contributed by vikkycirus

Javascript

function occurredOnce(arr, n) {
// counting frequency of every element
// using Map
let mp = new Map();
for(let i=0; i<n; i++) {
mp.set(arr[i], (mp.get(arr[i]) || 0) + 1);
}

// Traverse the map and print all
// the elements with occurrence 1
for(let [key, value] of mp) {
if(value == 1) {
console.log(key + " ");
}
}
}

// Driver code
let arr = [-9, -8, 4, 4, 5, 5, -1];
let n = arr.length;
occurredOnce(arr, n);

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;

class Program {

    // Function to find the elements that
    // appeared only once in the array
    static void OccurredOnce(int[] arr, int n) {
        // counting frequency of every element using .Count() method
        Dictionary<int, int> mp = arr.GroupBy(x => x).ToDictionary(g => g.Key, g => g.Count());

        // Traverse the map and print all
        // the elements with occurrence 1
        foreach (var item in mp) {
            if (item.Value == 1) {
                Console.Write(item.Key + " ");
            }
        }
    }

    // Driver code
    static void Main(string[] args) {
        int[] arr = {7, 7, 8, 8, 9, 1, 1, 4, 2, 2};
        int n = arr.Length;
        OccurredOnce(arr, n);
    }
}

// This code is contributed by Prince Kumar

Output

9 4 

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