LeetCode-in-Java

2284. Sender With Largest Word Count

Medium

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:

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:

Solution

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public String largestWordCount(String[] messages, String[] senders) {
        HashMap<String, Integer> x = new HashMap<>();
        for (int i = 0; i < messages.length; i++) {
            int words = messages[i].length() - messages[i].replace(" ", "").length() + 1;
            if (x.containsKey(senders[i])) {
                x.put(senders[i], x.get(senders[i]) + words);
            } else {
                x.put(senders[i], words);
            }
        }
        String result = "";
        int max = 0;
        for (Map.Entry<String, Integer> entry : x.entrySet()) {
            if (entry.getValue() > max
                    || entry.getValue() == max && result.compareTo(entry.getKey()) < 0) {
                max = entry.getValue();
                result = entry.getKey();
            }
        }
        return result;
    }
}