EASY || SIMPLE || BASIC
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n\n    int helper(string &s){\n        int ans = 0;\n        int i=0;\n        while(i < s.length()){\n            while(i < s.length() && s[i] != \' \'){\n                i++;\n            }\n            i++;\n            ans++;\n        }\n        return ans;\n    }\n    string largestWordCount(vector<string>& messages, vector<string>& senders) {\n        unordered_map<string, int> mp;\n        string ans = "";\n        int mx = 0;\n\n        for(int i=0;i<messages.size();i++){\n            int count = helper(messages[i]);\n            mp[senders[i]] += count;\n            if(mp[senders[i]] == mx){\n               ans = max(ans ,senders[i] );\n            }\n            else if(mp[senders[i]] > mx){\n                mx = mp[senders[i]];\n                ans = senders[i];\n            }\n        }\n        return ans;\n    }\n};\n```
['C++']