2284. Sender With Largest Word Count

Problem

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 <= 10^4

• 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 (Java)

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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).