Simple Space count solution || No String-Stream
\nclass Solution {\npublic:\n    string largestWordCount(vector<string>& messages, vector<string>& senders) {\n        map<string,int> m;\n        int n=senders
```\nclass Solution {\npublic:\n    string largestWordCount(vector<string>& messages, vector<string>& senders) {\n        map<string,int> m;\n        int n=senders.size();\n        for(int i=0;i<n;i++)\n        {\n            int cnt=0;\n            for(int j=0;j<messages[i].size();j++)\n            {\n                if(messages[i][j]==\' \')cnt++; //counting spaces\n            }\n            m[senders[i]]+=(cnt+1);//words count = space count+1 bcz of no trailing and leading spaces\n        }\n        vector<pair<int,string>> v;\n        for(auto& it:m)\n        {\n            v.push_back({it.second,it.first}); //sorting according to words sent\n        }\n        sort(v.begin(),v.end());\n        return v[v.size()-1].second; //even if there is a tie lexicographically larger will be at last\n    }\n};\n```
['C']