# 359  Logger Rate Limiter

## 359. [Logger Rate Limiter](https://leetcode.com/problems/logger-rate-limiter/description/)

## 1. Question

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;
```

## 2. Implementation

**(1) HashMap**

```java
class Logger {
    Map<String, Integer> map;

    /** Initialize your data structure here. */
    public Logger() {
        map = new HashMap<>();
    }

    /** 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) {
        if (!map.containsKey(message) || (timestamp - map.get(message)) >= 10) {
            map.put(message, timestamp);
            return true;
        }
        return false;
    }
}

/**
 * Your Logger object will be instantiated and called as such:
 * Logger obj = new Logger();
 * boolean param_1 = obj.shouldPrintMessage(timestamp,message);
 */
```

**(2) Set + Deque**

思路: 上一种方法由于不停将message存在hashmap中，所以空间消耗很大。为了优化空间，我们采用Deque，当在deque的头比timestamp小于10秒之内时，我们将其从deque remove

```java
class Logger {
    class Log {
        int time;
        String message;

        public Log(int time, String message) {
            this.time = time;
            this.message = message;
        }
    }

    private Set<String> set;
    private Deque<Log> deque;

    /** Initialize your data structure here. */
    public Logger() {
        set = new HashSet();
        deque = new ArrayDeque();
    }

    /** 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 (!deque.isEmpty() && (deque.peek().time <= timestamp - 10)) {
            Log oldLog = deque.remove();
            set.remove(oldLog.message);
        }

        if (!set.contains(message)) {
            set.add(message);
            deque.add(new Log(timestamp, message));
            return true;
        }
        return false;
    }
}

/**
 * Your Logger object will be instantiated and called as such:
 * Logger obj = new Logger();
 * boolean param_1 = obj.shouldPrintMessage(timestamp,message);
 */
```

## 3. Time & Space Complexity

**HashMap:** 时间复杂度O(1), 空间复杂度O(n)

**Set + Deque:** 时间复杂度O(1), 空间复杂度O(n)


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