# 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.

Example:
I/P = [1, 2, 3, 2, 3, 1, 3]
O/P = 3

## 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).

Program :

```// Java program to find the element occurring
// odd number of times
class OddOccurrence {

// funtion 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
```

Output :

`5`

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 :

```//Java program to find the element occurring odd
// number of times
import java.io.*;
import java.util.HashMap;

class OddOccurrence
{
// funtion 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
```

Output :

`5`

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 are implementations of this best approach.

Program:

## C/C++

```
//C program to find the element occurring odd number of times

#include <stdio.h>
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;
}

/* Diver 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]);
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
```

## Python

```
# 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)
```

Output :

`5`

Time Complexity:
O(n)

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