[Python 3] 2 solutions, one pass and dictionary search
Find the max sender name durion one iteration:\npython3 []\nclass Solution:\n    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:\n  
##### Find the max sender name durion one iteration:\n```python3 []\nclass Solution:\n    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:\n        d, maxSenderName= defaultdict(int), \'\'\n        for mes, name in zip(messages, senders):\n            d[name] += len(mes.split())\n            if d[name] > d[maxSenderName]:\n                maxSenderName = name\n            elif d[name] == d[maxSenderName] and name > maxSenderName:\n                maxSenderName = name\n        \n        return maxSenderName\n```\n##### Do search by dictionary:\n```python3 []\nclass Solution:\n    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:\n        d = defaultdict(int)\n        for m, n in zip(messages, senders):\n            d[n] += len(m.split())\n        return max(d, key = lambda k: (d[k], k))\n```
['Hash Table', 'Python', 'Python3']