Design a logger system that receive stream of messages along with its timestamps, each message should be printed if and only if it is not printed in the last 10 seconds.
Given a message and a timestamp (in seconds granularity), return true if the message should be printed in the given timestamp, otherwise returns false.
It is possible that several messages arrive roughly at the same time.
Example:

Logger logger = new Logger();

// logging string "foo" at timestamp 1
logger.shouldPrintMessage(1, "foo"); returns true; 

// logging string "bar" at timestamp 2
logger.shouldPrintMessage(2,"bar"); returns true;

// logging string "foo" at timestamp 3
logger.shouldPrintMessage(3,"foo"); returns false;

// logging string "bar" at timestamp 8
logger.shouldPrintMessage(8,"bar"); returns false;

// logging string "foo" at timestamp 10
logger.shouldPrintMessage(10,"foo"); returns false;

// logging string "foo" at timestamp 11
logger.shouldPrintMessage(11,"foo"); returns true;

Solution1：Hashmap

思路： map(message, vali_timestamp)，vali_timestamp是表示 出現過狀態的最晚時間
Time Complexity: O(N) Space Complexity: O(N)

Solution2：space friendly [design] 小(舊)頂堆 過濾 保持最大的10

思路： map(message, vali_timestamp)，vali_timestamp是表示 出現過狀態的最晚時間
Time Complexity: O() Space Complexity: O(N)

Solution1 Code：

class Logger {
    private Map<String, Integer> vali_map;
    
    public Logger() {
        vali_map = new HashMap<>();;
    }
    
    public boolean shouldPrintMessage(int timestamp, String message) {
        if (timestamp < vali_map.getOrDefault(message, 0))
            return false;
        ok.put(message, timestamp + 10);
        return true;
    }
}

Solution2 Code：

public class Logger {
    class Log {
        int timestamp;
        String message;
        public Log(int timestamp, String message) {
            this.timestamp = timestamp;
            this.message = message;
        }
    }
    
    PriorityQueue<Log> recent_logs;
    Set<String> recent_messages;   
    
    /** Initialize your data structure here. */
    public Logger() {
        recent_logs = new PriorityQueue<Log>(10, new Comparator<Log>() {
            public int compare(Log l1, Log l2) {
                return l1.timestamp - l2.timestamp;
            }
        });
        
        recent_messages = new HashSet<String>();
    }
    
    /** Returns true if the message should be printed in the given timestamp, otherwise returns false.
        If this method returns false, the message will not be printed.
        The timestamp is in seconds granularity. */
    public boolean shouldPrintMessage(int timestamp, String message) {
        while (recent_logs.size() > 0)   {
            Log log = recent_logs.peek();
            // discard the logs older than 10 minutes
            if (timestamp - log.timestamp >= 10) {
                recent_logs.poll();
                recent_messages.remove(log.message);
            } 
            else { 
                break;
            }
        }
        boolean res = !recent_messages.contains(message);
        if (res) {
            recent_logs.add(new Log(timestamp, message));
            recent_messages.add(message);
        }
        return res;
    }
}