Find the Number Occurring Odd Number of Times

Difficulty Level : Easy
Last Updated : 07 Apr, 2021

Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time & constant space.
Examples :

Input : arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5

A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and inner loop counts number of occurrences of the element picked by outer loop. Time complexity of this solution is O(n²).
Below is the implementation of the brute force approach :

C++

// C++ program to find the element
// occurring odd number of times
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the element
// occurring odd number of times
int getOddOccurrence(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;
}
 
// driver code
int main()
    {
        int arr[] = { 2, 3, 5, 4, 5, 2,
                      4, 3, 5, 2, 4, 4, 2 };
        int n = sizeof(arr) / sizeof(arr[0]);
 
        // Function calling
        cout << getOddOccurrence(arr, n);
 
        return 0;
    }

Java

// Java program to find the element occurring
// odd number of times
class OddOccurrence {
     
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int arr[], int arr_size)
    {
        int i;
        for (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;
    }
     
    // driver code
    public static void main(String[] args)
    {
        int arr[] = new int[]{ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.length;
        System.out.println(getOddOccurrence(arr, n));
    }
}
// This code has been contributed by Kamal Rawal

Python3

# Python program to find the element occurring
# odd number of times
     
# 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
     
     
# driver code
arr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]
n = len(arr)
print(getOddOccurrence(arr, n))
 
# This code has been contributed by
# Smitha Dinesh Semwal

C#

// C# program to find the element
// occurring odd number of times
using System;
 
class GFG
{
    // Function to find the element
    // occurring odd number of times
    static int getOddOccurrence(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;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.Length;
        Console.Write(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to find the
// element occurring odd
// number of times
 
// 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;
}
 
// Driver code
$arr = array(2, 3, 5, 4, 5, 2,
             4, 3, 5, 2, 4, 4, 2);
$n = sizeof($arr);
 
// Function calling
echo(getOddOccurrence($arr, $n));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
 
// Javascript program to find the element
// occurring odd number of times
 
// 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;
 
        // Function calling
        document.write(getOddOccurrence(arr, n));
 
// This code is contributed by Mayank Tyagi
 
</script>

Output :

A Better Solution is to use Hashing. Use array elements as key and their counts as value. Create an empty hash table. One by one traverse the given array elements and store counts. Time complexity of this solution is O(n). But it requires extra space for hashing.
Program :

C++

// C++ program to find the element 
// occurring odd number of times
#include <bits/stdc++.h>
using namespace std;
 
// function to find the element
// occurring odd number of times
int getOddOccurrence(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;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 4, 5, 2, 4,
                    3, 5, 2, 4, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function calling
    cout << getOddOccurrence(arr, n);
 
    return 0;
}
 
// This code is contributed by codeMan_d.

Java

// Java program to find the element occurring odd
// number of times
import java.io.*;
import java.util.HashMap;
 
class OddOccurrence
{
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int arr[], int n)
    {
        HashMap<Integer,Integer> hmap = new HashMap<>();
         
        // Putting all elements into the HashMap
        for(int i = 0; i < n; i++)
        {
            if(hmap.containsKey(arr[i]))
            {
                int val = hmap.get(arr[i]);
                         
                // If array element is already present then
                // increase the count of that element.
                hmap.put(arr[i], val + 1);
            }
            else
                 
                // if array element is not present then put
                // element into the HashMap and initialize
                // the count to one.
                hmap.put(arr[i], 1);
        }
 
        // Checking for odd occurrence of each element present
        // in the HashMap
        for(Integer a:hmap.keySet())
        {
            if(hmap.get(a) % 2 != 0)
                return a;
        }
        return -1;
    }
         
    // driver code   
    public static void main(String[] args)
    {
        int arr[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = arr.length;
        System.out.println(getOddOccurrence(arr, n));
    }
}
// This code is contributed by Kamal Rawal

Python3

# Python3 program to find the element 
# occurring odd number of times
  
# 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
 
  
# Driver code
arr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2]
n = len(arr)
  
# Function calling
print(getOddOccurrence(arr, n))
 
# This code is contributed by mohit kumar

C#

// C# program to find the element occurring odd
// number of times
using System;
using System.Collections.Generic;
 
public class OddOccurrence
{
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int []arr, int n)
    {
        Dictionary<int,int> hmap = new Dictionary<int,int>();
         
        // Putting all elements into the HashMap
        for(int i = 0; i < n; i++)
        {
            if(hmap.ContainsKey(arr[i]))
            {
                int val = hmap[arr[i]];
                         
                // If array element is already present then
                // increase the count of that element.
                hmap.Remove(arr[i]);
                hmap.Add(arr[i], val + 1);
            }
            else
                 
                // if array element is not present then put
                // element into the HashMap and initialize
                // the count to one.
                hmap.Add(arr[i], 1);
        }
 
        // Checking for odd occurrence of each element present
        // in the HashMap
        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)
    {
        int []arr = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = arr.Length;
        Console.WriteLine(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Princi Singh

Output :

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring element. Please note that XOR of two elements is 0 if both elements are same and XOR of a number x with 0 is x.
Below is the implementation of the above approach.

C++

// C++ program to find the element
// occurring odd number of times
#include <bits/stdc++.h>
using namespace std;
 
// 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;
}
 
/* Driver function to test above function */
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
    cout << getOddOccurrence(ar, n);
     
    return 0;
}

C

// C program to find the element
// occurring odd number of times
#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;
}
 
/* Driver function to test above function */
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;
}

Java

//Java program to find the element occurring odd number of times
 
class OddOccurance
{
    int getOddOccurrence(int ar[], int ar_size)
    {
        int i;
        int res = 0;
        for (i = 0; i < ar_size; i++)
        {
            res = res ^ ar[i];
        }
        return res;
    }
 
    public static void main(String[] args)
    {
        OddOccurance occur = new OddOccurance();
        int ar[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = ar.length;
        System.out.println(occur.getOddOccurrence(ar, n));
    }
}
// This code has been contributed by Mayank Jaiswal

Python3

# 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))

C#

// C# program to find the element
// occurring odd number of times
using System;
 
class GFG
{
    // Function to find the element
    // occurring odd number of times
    static int getOddOccurrence(int []arr, int arr_size)
    {
        int res = 0;
        for (int i = 0; i < arr_size; i++)
        {
            res = res ^ arr[i];
        }
        return res;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.Length;
        Console.Write(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to find the
// element occurring odd
// number of times
 
// 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));
     
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>
 
// JavaScript program to find the element
// occurring odd number of times
 
// Function to find the element
// occurring odd number of times
function getOddOccurrence( ar, ar_size)
{
    let res = 0;
     
    for (let i = 0; i < ar_size; i++)    
        res = res ^ ar[i];
     
    return res;
}
 
    // driver code
 
    let arr = [ 2, 3, 5, 4, 5, 2, 4,
                3, 5, 2, 4, 4, 2 ];
    let n = arr.length;
 
    // Function calling
    document.write(getOddOccurrence(arr, n));
 
 
</script>

Output :

Time Complexity: O(n)

Method 3:Using Built in python functions:

Count the frequencies of every element using Counter function
Traverse in frequency dictionary
Check which element has a odd frequency.
Print that element and break the loop

Below is the implementation:

Python3

# importing counter from colleections
from collections import Counter
 
# Python3 implementation to find
# odd frequeency element
def oddElement(arr, n):
 
    # Calculating freequencies usinf Counter
    count_map = Counter(arr)
 
    for i in range(0, n):
 
        # If count of element is odd we return
        if (count_map[arr[i]] % 2 != 0):
            return arr[i]
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 1, 3, 3, 5, 6, 6]
    n = len(arr)
    print(oddElement(arr, n))
 
# This code is contributed by vikkycirus

Output:

Time Complexity: O(N)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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