[C++/python] one-pass and 2-line python solution
Approach 1: one-pass\nTime/Space Complexity: O(N1+N2); where N1 and N2 are the length of s1 and s2, respectively\n\nclass Solution {\npublic:\n    vector<string
Approach 1: one-pass\nTime/Space Complexity: O(N1+N2); where N1 and N2 are the length of s1 and s2, respectively\n```\nclass Solution {\npublic:\n    vector<string> uncommonFromSentences(string s1, string s2) {\n        stringstream ss(s1 + " " + s2);        \n        unordered_map<string, int> m;\n        for(string word;ss>>word;)  ++m[word];\n        vector<string> ans;\n        for(auto& [k, v]: m)\n            if(v == 1) ans.emplace_back(k);\n        return ans;\n    }\n};\n```\n\nApproach 2: Python\n```\nclass Solution:\n    def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:\n        count = collections.Counter((s1 + \' \' + s2).split())\n        #return [k for k,v in count.items() if v == 1] # ok\n        return [word for word in count if count[word] == 1]    \n    \n```
['C', 'Python']