Elements that occurred only once in the array - Coding

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:

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

<?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++

<?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)

Reffered: https://www.geeksforgeeks.org

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