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Sender With Largest Word Count

You have a chat log of n messages. You are given two string arrays messages and senders where messages[i] is a message sent by senders[i].

A message is list of words that are separated by a single space with no leading or trailing spaces. The word count of a sender is the total number of words sent by the sender. Note that a sender may send more than one message.

Return the sender with the largest word count. If there is more than one sender with the largest word count, return the one with the lexicographically largest name.

Note:

  • Uppercase letters come before lowercase letters in lexicographical order.
  • "Alice" and "alice" are distinct.

 

Example 1:

Input: messages = ["Hello userTwooo","Hi userThree","Wonderful day Alice","Nice day userThree"], senders = ["Alice","userTwo","userThree","Alice"]
Output: "Alice"
Explanation: Alice sends a total of 2 + 3 = 5 words.
userTwo sends a total of 2 words.
userThree sends a total of 3 words.
Since Alice has the largest word count, we return "Alice".

Example 2:

Input: messages = ["How is leetcode for everyone","Leetcode is useful for practice"], senders = ["Bob","Charlie"]
Output: "Charlie"
Explanation: Bob sends a total of 5 words.
Charlie sends a total of 5 words.
Since there is a tie for the largest word count, we return the sender with the lexicographically larger name, Charlie.

 

Constraints:

  • n == messages.length == senders.length
  • 1 <= n <= 104
  • 1 <= messages[i].length <= 100
  • 1 <= senders[i].length <= 10
  • messages[i] consists of uppercase and lowercase English letters and ' '.
  • All the words in messages[i] are separated by a single space.
  • messages[i] does not have leading or trailing spaces.
  • senders[i] consists of uppercase and lowercase English letters only.

Solution Explanation for Sender With Largest Word Count

This problem requires finding the sender with the largest word count in a series of messages. If there's a tie, the lexicographically largest sender name should be returned.

Approach:

The most efficient approach uses a hash table (dictionary in Python) to store the word count for each sender. We iterate through the messages and senders arrays simultaneously. For each message, we count the words (by counting spaces plus one) and add it to the sender's count in the hash table. After processing all messages, we iterate through the hash table to find the sender with the maximum word count. If multiple senders have the same maximum count, we select the one with the lexicographically larger name.

Time Complexity Analysis:

  • Counting words: Iterating through each message to count words takes O(L), where L is the total length of all messages. This step is done once for each message, resulting in a total time of O(nL) where n is the number of messages. In the optimized code (using count() in C++, split() in TypeScript, or strings.Count() in Go), the time complexity becomes O(L) in total as we can count spaces efficiently.
  • Iterating through the hash table: Iterating through the hash table to find the maximum word count takes O(n) time in the worst case (where n is the number of unique senders). In the best case, it would be O(1) if we already find the max count in the first iteration. This is a relatively small operation compared to the word counting part.
  • Overall: The overall time complexity is dominated by counting words in the messages, which is O(L) with optimized word counting. However, in the worst case with inefficient word counting, it will be O(nL).

Space Complexity Analysis:

The space complexity is determined by the size of the hash table. In the worst case, each sender has a unique entry, leading to O(n) space complexity where n is the number of senders. This is because we store the word counts for each sender.

Code Implementation (Python):

from collections import Counter
 
class Solution:
    def largestWordCount(self, messages: List[str], senders: List[str]) -> str:
        word_counts = Counter()  #Efficient counter for word counts
        for message, sender in zip(messages, senders):
            word_counts[sender] += message.count(" ") + 1 #Efficient word count
        max_count = 0
        result_sender = ""
        for sender, count in word_counts.items():
            if count > max_count:
                max_count = count
                result_sender = sender
            elif count == max_count and sender > result_sender:  #Lexicographical comparison
                result_sender = sender
        return result_sender

The code in other languages (Java, C++, Go, TypeScript) follow the same logic, only differing in syntax and data structures. The optimized versions using built-in functions for counting words significantly reduce the time complexity from O(nL) to O(L) for word counting. The overall complexity is then O(L) where L is the total number of characters in the messages plus O(n) for maintaining the word counts in the hash table. Therefore, overall, the time complexity is O(L) dominated by the word counting. The space complexity remains O(n).