# 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

``````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%)

``````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 {
}
}
if (numbers.size() == 1) {
Iterator<Integer> iter = numbers.iterator();
return (int) iter.next();
}
return 0;
}
}
``````

Hash Set - Similar solution (8ms)

``````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{
}
}
Iterator<Integer> itr = set.iterator();
return itr.next();
}
}
``````

## Reference

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