Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
'e' is the letter that was added.

My original solution: using HashMap to store the count of appearance of letters in string s, and check if there is match in string t:

class Solution {
    public char findTheDifference(String s, String t) {
        Map<Character,Integer> map = new HashMap<>();
        for(int i=0;i<s.length();i++){
            int count=map.getOrDefault(s.charAt(i),0)+1;
            map.put(s.charAt(i),count);
        }

        for(int i=0;i<t.length();i++){
            if(!map.containsKey(t.charAt(i))) return t.charAt(i);
            else{
                int count = map.get(t.charAt(i));
                if(count<1) return t.charAt(i);

                map.put(t.charAt(i),--count);
            }
        }
        return 0;
    }
}

However, there is a much smarter way to use XOR.
Since 0^N=N, N^N=0,
So if string s and t is ever the same, the XOR of each letter in the 2 strings should be zero.
The leftover character should be the excessive one.

public char findTheDifference(String s, String t) {
    char c = 0;
    for (int i = 0; i < s.length(); ++i) {
        c ^= s.charAt(i);
    }
    for (int i = 0; i < t.length(); ++i) {
        c ^= t.charAt(i);
    }
    return c;
}