we use bitwise `XOR`

to solve this problem :

0 ^ N = N

N ^ N = 0

So..... if N is the added number

N1 ^ N1 ^ N2 ^ N2 ^..............^ Nx ^ Nx ^ N

= (N1^N1) ^ (N2^N2) ^..............^ (Nx^Nx) ^ N

= 0 ^ 0 ^ ..........^ 0 ^ N

= N

```
public char findTheDifference(String s, String t) {
int result = 0;
for (char c : s.toCharArray() ) {result = result ^ c;}
for (char c : t.toCharArray() ) {result = result ^ c;}
return (char)(result);
}
```