Given an array of integers. All numbers occur twice except one number which occurs once. Find the number in O(n) time & constant extra space.

**Example :**

Input: ar[] = {7, 3, 5, 4, 5, 3, 4} Output: 7

One solution is to check every element if it appears once or not. Once an element with a single occurrence is found, return it. Time complexity of this solution is O(n2).

A better solution is to use hashing.

- Traverse all elements and put them in a hash table. Element is used as key and the count of occurrences is used as the value in the hash table.
- Traverse the array again and print the element with count 1 in the hash table.

This solution works in O(n) time but requires extra space.

The best solution is to use XOR. XOR of all array elements gives us the number with a single occurrence. The idea is based on the following two facts.

a) XOR of a number with itself is 0.

b) XOR of a number with 0 is number itself.

Let us consider the above example. Let ^ be xor operator as in C and C++. res = 7 ^ 3 ^ 5 ^ 4 ^ 5 ^ 3 ^ 4 Since XOR is associative and commutative, above expression can be written as: res = 7 ^ (3 ^ 3) ^ (4 ^ 4) ^ (5 ^ 5) = 7 ^ 0 ^ 0 ^ 0 = 7 ^ 0 = 7

Below are implementations of above algorithm.

`<script>`

`// JavScript program to find the array`

`// element that appears only once`

`function`

`findSingle(ar, ar_size)`

` `

`{`

` `

`// Do XOR of all elements and return`

` `

`let res = ar[0];`

` `

`for`

`(let i = 1; i < ar_size; i++)`

` `

`res = res ^ ar[i];`

` `

`return`

`res;`

` `

`}`

`// Driver code `

` `

`let ar = [2, 3, 5, 4, 5, 3, 4];`

` `

`let n = ar.length;`

` `

`document.write(`

`"Element occurring once is "`

` `

`+ findSingle(ar, n));`

` `

`</script>`

**Output:**

Element occurring once is 2

The time complexity of this solution is O(n) and it requires O(1) extra space.