Number occurring odd number of times
Last Updated :
30 Aug, 2025
Given an array arr[] of positive integers. All numbers occur an even number of times except one number which occurs an odd number of times.
Examples :
Input : arr[]= [1, 2, 3, 2, 3, 1, 3]
Output : 3
Input : arr[] = [5, 7, 2, 7, 5, 2, 5]
Output : 5
[Naive Approach] Using Nested Loop
A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop.
C++
#include<iostream>
#include <vector>
using namespace std;
// Function to find the element
// occurring odd number of times
int getOddOccurrence(vector<int>&arr, int n)
{
for (int i = 0; i < n ; i++) {
int count = 0;
for (int j = 0; j < n; j++)
{
if (arr[i] == arr[j])
count++;
}
if (count % 2 != 0)
return arr[i];
}
return -1;
}
int main()
{
vector<int>arr { 2, 3, 5, 4, 5, 2,
4, 3, 5, 2, 4, 4, 2 };
int n = arr.size();
cout << getOddOccurrence(arr, n);
return 0;
}
import java.util.*;
public class Main {
// Function to find the element
// occurring odd number of times
static int getOddOccurrence(ArrayList<Integer> arr, int n) {
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr.get(i).equals(arr.get(j)))
count++;
}
if (count % 2 != 0)
return arr.get(i);
}
return -1;
}
public static void main(String[] args) {
ArrayList<Integer> arr = new
ArrayList<>(Arrays.asList
(2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2));
int n = arr.size();
System.out.println(getOddOccurrence(arr, n));
}
}
# function to find the element occurring odd
# number of times
def getOddOccurrence(arr, arr_size):
for i in range(0,arr_size):
count = 0
for j in range(0, arr_size):
if arr[i] == arr[j]:
count+=1
if (count % 2 != 0):
return arr[i]
return -1
if __name__ == "__main__":
arr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]
n = len(arr)
print(getOddOccurrence(arr, n))
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the element
// occurring odd number of times
static int GetOddOccurrence(List<int> arr, int arr_size)
{
for (int i = 0; i < arr_size; i++)
{
int count = 0;
for (int j = 0; j < arr_size; j++)
{
if (arr[i] == arr[j])
count++;
}
if (count % 2 != 0)
return arr[i];
}
return -1;
}
public static void Main()
{
List<int> arr = new List<int> { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
int n = arr.Count;
Console.Write(GetOddOccurrence(arr, n));
}
}
// Function to find the element
// occurring odd number of times
function getOddOccurrence(arr, arr_size)
{
for (let i = 0; i < arr_size; i++) {
let count = 0;
for (let j = 0; j < arr_size; j++)
{
if (arr[i] == arr[j])
count++;
}
if (count % 2 != 0)
return arr[i];
}
return -1;
}
//Driver Code
let arr = [ 2, 3, 5, 4, 5, 2,
4, 3, 5, 2, 4, 4, 2 ];
let n = arr.length;
console.log(getOddOccurrence(arr, n));
<?php
// Function to find the element
// occurring odd number of times
function getOddOccurrence(&$arr, $arr_size)
{
$count = 0;
for ($i = 0;
$i < $arr_size; $i++)
{
for ($j = 0;
$j < $arr_size; $j++)
{
if ($arr[$i] == $arr[$j])
$count++;
}
if ($count % 2 != 0)
return $arr[$i];
}
return -1;
}
$arr = array(2, 3, 5, 4, 5, 2,
4, 3, 5, 2, 4, 4, 2);
$n = sizeof($arr);
echo(getOddOccurrence($arr, $n));
?>
Time Complexity: O(n^2)
Auxiliary Space: O(1)
[Another Approach]Using Hashing
A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts.
C++
#include <iostream>
#include<vector>
#include <unordered_map>
using namespace std;
// function to find the element
// occurring odd number of times
int getOddOccurrence(vector<int>&arr,int size)
{
// Defining HashMap in C++
unordered_map<int, int> hash;
// Putting all elements into the HashMap
for(int i = 0; i < size; i++)
{
hash[arr[i]]++;
}
// Iterate through HashMap to check an element
// occurring odd number of times and return it
for(auto i : hash)
{
if(i.second % 2 != 0)
{
return i.first;
}
}
return -1;
}
int main()
{
vector<int>arr { 2, 3, 5, 4, 5, 2, 4,
3, 5, 2, 4, 4, 2 };
int n = arr.size();
cout << getOddOccurrence(arr, n);
return 0;
}
import java.util.*;
class OddOccurrence {
// Function to find the element
// occurring odd number of times
static int getOddOccurrence(ArrayList<Integer> arr, int n) {
for (int i = 0; i < n; i++) {
int count = 0;
// Count occurrences of arr.get(i)
for (int j = 0; j < n; j++) {
if (arr.get(i).equals(arr.get(j)))
count++;
}
// If count is odd, return it
if (count % 2 != 0)
return arr.get(i);
}
return -1; // No odd occurrence found
}
public static void main(String[] args) {
ArrayList<Integer> arr =
new ArrayList<>(Arrays.asList(2, 3, 5, 4, 5, 2,
4, 3, 5, 2, 4, 4, 2));
int n = arr.size();
System.out.println(getOddOccurrence(arr, n));
}
}
# function to find the element
# occurring odd number of times
def getOddOccurrence(arr,size):
# Defining HashMap in C++
Hash=dict()
# Putting all elements into the HashMap
for i in range(size):
Hash[arr[i]]=Hash.get(arr[i],0) + 1;
# Iterate through HashMap to check an element
# occurring odd number of times and return it
for i in Hash:
if(Hash[i]% 2 != 0):
return i
return -1
if __name__ == "__main__":
arr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2]
n = len(arr)
print(getOddOccurrence(arr, n))
using System;
using System.Collections.Generic;
public class GFG
{
// Function to find the element
// occurring odd number of times
static int GetOddOccurrence(List<int> arr, int n)
{
Dictionary<int, int> hmap = new Dictionary<int, int>();
// Putting all elements into the Dictionary
for (int i = 0; i < n; i++)
{
if (hmap.ContainsKey(arr[i]))
{
int val = hmap[arr[i]];
// If element is already present then
// increase its count
hmap[arr[i]] = val + 1;
}
else
{
// If element is not present then add
// it with count = 1
hmap.Add(arr[i], 1);
}
}
// Checking for odd occurrence of each
// element present in the Dictionary
foreach (KeyValuePair<int, int> entry in hmap)
{
if (entry.Value % 2 != 0)
{
return entry.Key;
}
}
return -1;
}
// Driver code
public static void Main(String[] args)
{
List<int> arr = new List<int>
{ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
int n = arr.Count;
Console.WriteLine(GetOddOccurrence(arr, n));
}
}
// function to find the element occurring odd
// number of times
function getOddOccurrence(arr,n)
{
let hmap = new Map();
// Putting all elements into the HashMap
for(let i = 0; i < n; i++)
{
if(hmap.has(arr[i]))
{
let val = hmap.get(arr[i]);
// If array element is already present then
// increase the count of that element.
hmap.set(arr[i], val + 1);
}
else
{
// if array element is not present then put
// element into the HashMap and initialize
// the count to one.
hmap.set(arr[i], 1);
}
}
// Checking for odd occurrence of
// each element present in the HashMap
for(let [key, value] of hmap.entries())
{
//console.log(hmap[a]+"<br>")
if(hmap.get(key) % 2 != 0)
return key;
}
return -1;
}
// Driver code
let arr=[2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2];
let n = arr.length;
console.log(getOddOccurrence(arr, n));
// This code is contributed by unknown2108
Time Complexity: O(n)
Auxiliary Space: O(n)
[Expected Approach] Using Bit Manipulation
The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements.
Here ^ is the XOR operators;
Note : x ^ 0 = x
x ^ y=y ^ x (Commutative property holds)
(x ^ y) ^ z = x ^ ( y ^ z) (Distributive property holds)
x ^ x=0
C++
#include <iostream>
#include <vector>
using namespace std;
// Function to find element occurring odd number of times
int getOddOccurrence(vector<int>& ar)
{
int res = 0;
for (int i = 0; i < ar.size(); i++)
res ^= ar[i];
return res;
}
int main()
{
vector<int> ar = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
// Function calling
cout << getOddOccurrence(ar);
return 0;
}
#include <stdio.h>
// Function to find element occurring
// odd number of times
int getOddOccurrence(int ar[], int ar_size)
{
int res = 0;
for (int i = 0; i < ar_size; i++)
res = res ^ ar[i];
return res;
}
int main()
{
int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
int n = sizeof(ar) / sizeof(ar[0]);
// Function calling
printf("%d", getOddOccurrence(ar, n));
return 0;
}
import java.util.*;
public class Main {
// Function to find element occurring odd number of times
static int getOddOccurrence(ArrayList<Integer> ar) {
int res = 0;
for (int i = 0; i < ar.size(); i++)
res ^= ar.get(i);
return res;
}
public static void main(String[] args) {
ArrayList<Integer> ar =
new ArrayList<>(Arrays.asList
(2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2));
// Function calling
System.out.println(getOddOccurrence(ar));
}
}
# Python program to find the element occurring odd number of times
def getOddOccurrence(arr):
# Initialize result
res = 0
# Traverse the array
for element in arr:
# XOR with the result
res = res ^ element
return res
# Test array
arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2]
print("%d" % getOddOccurrence(arr))
using System;
using System.Collections.Generic;
class GFG
{
// Function to find element occurring odd number of times
static int GetOddOccurrence(List<int> ar)
{
int res = 0;
for (int i = 0; i < ar.Count; i++)
res ^= ar[i];
return res;
}
static void Main()
{
List<int> ar = new List<int> { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
// Function calling
Console.WriteLine(GetOddOccurrence(ar));
}
}
function getOddOccurrence(arr) {
let res = 0;
for (let i = 0; i < arr.length; i++) {
res ^= arr[i];
}
return res;
}
// Driver code
let arr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2];
console.log(getOddOccurrence(arr));
<?php
// Function to find element
// occurring odd number of times
function getOddOccurrence(&$ar, $ar_size)
{
$res = 0;
for ($i = 0; $i < $ar_size; $i++)
$res = $res ^ $ar[$i];
return $res;
}
// Driver Code
$ar = array(2, 3, 5, 4, 5, 2,
4, 3, 5, 2, 4, 4, 2);
$n = sizeof($ar);
// Function calling
echo(getOddOccurrence($ar, $n));
?>
Time Complexity: O(n)
Auxiliary Space: O(1)
Exceptionally odd | DSA Problem
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