## Saturday, June 18, 2016

### LeetCode - 359 Logger Rate Limiter

http://www.cnblogs.com/grandyang/p/5592635.html
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;

http://dartmooryao.blogspot.com/2016/06/leetcode-359-logger-rate-limiter.html
Use a map to store the message and timestamp.
When the "shouldPrintMsg" is called, we check whether the map don't have the string or the previous timestamp is 10 seconds before.
If yes, we update the map with the new record and return true.
Otherwise, do nothing and return false
Map<String, Integer> msg;
public Logger() {
msg = 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(!msg.containsKey(message) || timestamp-msg.get(message)>=10){
msg.put(message, timestamp);
return true;
}else{
return false;
}
}
https://discuss.leetcode.com/topic/48359/short-c-java-python-bit-different/
https://discuss.leetcode.com/topic/52146/java-no-brainer-using-hashmap
``````    private Map<String, Integer> ok = new HashMap<>();

public boolean shouldPrintMessage(int timestamp, String message) {
if (timestamp < ok.getOrDefault(message, 0))
return false;
ok.put(message, timestamp + 10);
return true;
}``````

```    bool shouldPrintMessage(int timestamp, string message) {
if (!m.count(message)) {
m[message] = timestamp;
return true;
}
if (timestamp - m[message] >= 10) {
m[message] = timestamp;
return true;
}
return false;
}
unordered_map<string, int> m;```
public boolean shouldPrintMessage(int timestamp, String message) { //update timestamp of the message if the message is coming in for the first time,or the last coming time is earlier than 10 seconds from now if(!map.containsKey(message)||timestamp-map.get(message)>=10){ map.put(message,timestamp); return true; } return false; }
https://leetcode.com/discuss/108703/short-c-java-python-bit-different
private Map<String, Integer> ok = new HashMap<>(); public boolean shouldPrintMessage(int timestamp, String message) { if (timestamp < ok.getOrDefault(message, 0)) return false; ok.put(message, timestamp + 10); return true; }

https://discuss.leetcode.com/topic/48399/java-concurrenthashmap-solution/
``````    ConcurrentHashMap<String, Integer> lastPrintTime;

/** Initialize your data structure here. */
public Logger() {
lastPrintTime = new ConcurrentHashMap<String, Integer>();
}

/** Returns true if the message should be printed in the given timestamp, otherwise returns false. The timestamp is in seconds granularity. */
public boolean shouldPrintMessage(int timestamp, String message) {
Integer last = lastPrintTime.get(message);

return last == null && lastPrintTime.putIfAbsent(message, timestamp) == null
|| last != null && timestamp - last >= 10 && lastPrintTime.replace(message, last, timestamp);

}``````
``````    public Map<String, Integer> map;
int lastSecond = 0;

/** Initialize your data structure here. */
public Logger() {
map = new java.util.LinkedHashMap<String, Integer>(100, 0.6f, true) {
protected boolean removeEldestEntry(Map.Entry<String, Integer> eldest) {
return lastSecond - eldest.getValue() > 10;
}
};
}

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

https://discuss.leetcode.com/topic/48615/a-solution-that-only-keeps-part-of-the-messages/
``````class Log {
int timestamp;
String message;
public Log(int aTimestamp, String aMessage) {
timestamp = aTimestamp;
message = aMessage;
}
}

public class Logger {
PriorityQueue<Log> recentLogs;
Set<String> recentMessages;

/** Initialize your data structure here. */
public Logger() {
recentLogs = new PriorityQueue<Log>(10, new Comparator<Log>() {
public int compare(Log l1, Log l2) {
return l1.timestamp - l2.timestamp;
}
});

recentMessages = 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 (recentLogs.size() > 0)   {
Log log = recentLogs.peek();
// discard the logs older than 10 minutes
if (timestamp - log.timestamp >= 10) {
recentLogs.poll();
recentMessages.remove(log.message);
} else
break;
}
boolean res = !recentMessages.contains(message);
if (res) {
}
return res;
}
}``````