Elements that occurred only once in the array
Last Updated :
11 Jul, 2025
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++
// 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 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 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
<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# 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 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
?>
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 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 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
<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# 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
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 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 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
<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# 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 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
?>
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
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 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
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# 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
Time Complexity: O(n)
Auxiliary Space: O(n)
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Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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