from collections import Counter

def string_diff(s: str, t: str) -> int:
    s_count = Counter(s)
    t_count = Counter(t)
    for ch in set(s+t):
        if abs(s_count[ch] - t_count[ch]) == 1:
            return ch
    return ''

#Challenge

given a string s and a string t with t being s shuffled with optionally one extra character added, return this extra character or the empty string if it doesn't exist.

#Solution

count the number of times each character exists in s and t then return the character for which the absolute difference between the two is 1.

#Test Cases

string_diff('foobar', 'barfoot') // 't'
string_diff('ide', 'idea') // 'a'
string_diff('coding', 'ingcod') // ''

#Potential Optimisations

Instead of using counter we can a single dictionary with occurences in s incrementing the number and occurences in t decrementing it. Such as first unique character.