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 occurred only once in the array.

Note: Duplicates appear side by side every time. Might be few numbers can occur one time and just assume this is a right rotating array (just say an array can rotate k times towards right). 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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].

Below is the implementation of 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 

Python 3

# 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 

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

Time Complexity: O(Nlogn)
Space Complexity: 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 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 

Output:

4 9

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

Method-3: Using given assumptions.
It is given that an array can be rotated any times and duplicates will appear side by side every time. So after rotating the first and last element will appear side by side.

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

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 

Python 3

# 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 

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

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

C++

Java

Python3

C++

Java

Python 3

C#

PHP

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