## Codility - OddOccurrencesInArray

A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

```
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
```

- the elements at indexes 0 and 2 have value 9
- the elements at indexes 1 and 3 have value 3
- the elements at indexes 4 and 6 have value 9
- the element at index 5 has value 7 and is unpaired

Write a function that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

```
A[0] = 9 A[1] = 3 A[2] = 9
A[3] = 3 A[4] = 9 A[5] = 7
A[6] = 9
```

the function should return 7, as explained in the example above.

Assume that:

- N is an odd integer within the range [1..1,000,000]
- each element of array A is an integer within the range [1..1,000,000,000]
- all but one of the values in A occur an even number of times

Complexity:

- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

### Solution

This problem looks very complicated in the begining, but actually it’s quite simple.

We need to find out which number is not paired, and that can easily be accomplished by using bitwise operation, more to the point **XOR** operator.

```
0 0 = 0
0 A = A
A 0 = A
A A = 0
```

From this table we can see that everytime we use a XOR operator on the number it self it will reset to zero, that’s what we need.

So the solution is iterating the whole array and doing a XOR on a variable that will hold the missing number.

```
func Solution(A []int) int {
var number int = 0;
for _, v := range A {
number ^= v
}
return number
}
```