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Formatted question description: https://leetcode.ca/all/2284.html
2284. Sender With Largest Word Count
- Difficulty: Medium.
- Related Topics: Array, Hash Table, String, Counting.
- Similar Questions: Top K Frequent Elements, Top K Frequent Words.
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:
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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 (Java, C++, Python)
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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; } } -
Todo -
class Solution: def largestWordCount(self, messages: List[str], senders: List[str]) -> str: cnt = Counter() for msg, sender in zip(messages, senders): cnt[sender] += msg.count(' ') + 1 ans = '' for sender, v in cnt.items(): if cnt[ans] < v or (cnt[ans] == v and ans < sender): ans = sender return ans ############ # 2284. Sender With Largest Word Count # https://leetcode.com/problems/sender-with-largest-word-count/ class Solution: def largestWordCount(self, messages: List[str], senders: List[str]) -> str: mp = defaultdict(int) for message, sender in zip(messages, senders): mp[sender] += len(message.split()) currCount = float('-inf') res = "" for sender, count in mp.items(): if count > currCount: currCount = count res = sender elif count == currCount: res = max(res, sender) return res
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).