2284. Sender With Largest Word Count

Explanation

To solve this problem, we can iterate through the messages and keep track of the word count for each sender using a HashMap. After calculating the word count for each sender, we can then find the sender with the largest word count. In case of a tie, we return the sender with the lexicographically largest name.

Create a HashMap to store the word count for each sender.
Iterate through the messages array and split each message into words.
Update the word count for the sender of the current message.
Find the sender with the largest word count. If there is a tie, return the sender with the lexicographically largest name.

View Leetcode Problem

import java.util.*;

class Solution {
    public String largestSender(String[] messages, String[] senders) {
        Map<String, Integer> wordCount = new HashMap<>();
        
        for (int i = 0; i < messages.length; i++) {
            String sender = senders[i];
            String[] words = messages[i].split(" ");
            wordCount.put(sender, wordCount.getOrDefault(sender, 0) + words.length);
        }
        
        String largestSender = "";
        int maxCount = 0;
        
        for (Map.Entry<String, Integer> entry : wordCount.entrySet()) {
            if (entry.getValue() > maxCount || (entry.getValue() == maxCount && entry.getKey().compareTo(largestSender) > 0)) {
                maxCount = entry.getValue();
                largestSender = entry.getKey();
            }
        }
        
        return largestSender;
    }
}

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Explanation

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