Unique and Easy Python Solution | Unique approach | Memory usage less than 99.91%
\nclass Solution:\n    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:\n        \n        dic = {}\n        name = 0\n        while 
```\nclass Solution:\n    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:\n        \n        dic = {}\n        name = 0\n        while senders:\n            print(senders[name] , messages[name].lstrip().rstrip())\n            \n            if senders[name] not in dic:\n                dic[senders[name]] = len(messages[name].lstrip().rstrip().split(\' \'))\n\n            else:\n                dic[senders[name]] += len(messages[name].lstrip().rstrip().split(\' \'))\n            \n            senders.pop(name)\n            messages.pop(name)\n                \n        \n        d = {}\n        for key , val in dic.items():\n            if val not in d:\n                d[val] = [key]\n                \n            else:\n                d[val].append(key)\n                \n            d[val].sort(reverse = True)\n                \n            \n        return d[max(d)][0]\n```
['Python']