Find the Number Occurring Odd Number of Times

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 
?> 

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 
?> 

Output :

Time Complexity: O(n)

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

C++

C

Java

Python3

C#

PHP

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