LeetCode: Find the difference

code.js

/*
The first solution I thought of was:
turn both strings into arrays
sort both strings (js native sort method)
loop through the arrays 
when the letters aren't equal return the letter in the t array 
or if they're all equal return the letter at the end 

But using the sort method means the time complexity at worst > O(n).

How can I make it O(n)? lets find a solution that doesn't involve sorting.
current solution is O(n) time complexity and (I think) O(n) space complexity. 
*/
var findTheDifference = function(s, t) {
    var sFreq = {};
    var tFreq = {};    
    s = s.split("");
    t = t.split("");    
    
    s.forEach(function(letter){
     sFreq[letter] = sFreq[letter] + 1 || 1;   
    });
    
    t.forEach(function(letter){
     tFreq[letter] = tFreq[letter] + 1 || 1;           
    });    

    for (var key in tFreq) {
        if (!(key in sFreq)) {
          return key;
        }else { //match
          if(tFreq[key] !== sFreq[key]) {
            return key;
          }
        }
    }
};

Assumed arguments provided to function were not equal and both would be supplied.