# Single Number

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

**Note:**

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

```
Input: [2,2,1]
Output: 1
```

Example 2:

```
Input: [4,1,2,1,2]
Output: 4
```

### Hash Table

* Time complexity :O(n \* 1) = O(n). Time complexity of`for`loop is `O(n)`. Time complexity of hash table operation add/remove is `O(1)`.
* Space complexity : `O(n)`. The space required by hash set equal to the number of elements in `nums`.

### Bit Manipulation

```
If we take XOR of zero and some bit, it will return that bit
a⊕0=a
If we take XOR of two same bits, it will return 0
a⊕a=0
a⊕b⊕a=(a⊕a)⊕b=0⊕b=b
```

## Solution

The Most Elegant Way - Bit Manipulation

```java
class Solution {
    public int singleNumber(int[] nums) {

        int a = 0;
        for (int n : nums) a ^= n;
        return a ;

    }
}
```

Hash Set - O(n) space, O(n) time (8ms 38.85%)

```java
class Solution {
    public int singleNumber(int[] nums) {
        Set<Integer> numbers = new HashSet<Integer>();
        for (int num : nums) {
            if (numbers.contains(num)) {
                numbers.remove(num);
            } else {
                numbers.add(num);
            }
        }
        if (numbers.size() == 1) {
            Iterator<Integer> iter = numbers.iterator();  
            return (int) iter.next();
        }
        return 0;
    }
}
```

Hash Set - Similar solution (8ms)

```java
class Solution {
    public int singleNumber(int[] nums) {
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++){
            if (set.contains(nums[i])){
                set.remove(nums[i]);
            }
            else{
                set.add(nums[i]);
            }
        }
        Iterator<Integer> itr = set.iterator();
        return itr.next();
    }
}
```

## Reference

<https://leetcode.com/articles/single-number/>