Design a hit counter which counts the number of hits received in the past 5 minutes.
Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the earliest timestamp starts at 1.
It is possible that several hits arrive roughly at the same time.
Example:
HitCounter counter = new HitCounter();
// hit at timestamp 1.
counter.hit(1);
// hit at timestamp 2.
counter.hit(2);
// hit at timestamp 3.
counter.hit(3);
// get hits at timestamp 4, should return 3.
counter.getHits(4);
// hit at timestamp 300.
counter.hit(300);
// get hits at timestamp 300, should return 4.
counter.getHits(300);
// get hits at timestamp 301, should return 3.
counter.getHits(301);
Follow up:
What if the number of hits per second could be very large? Does your design scale?
更进一步的，是因为已知时间窗口，可以设定若干bucket，比如60 spots for 60 seconds。这样每一秒的hit都能归入一个bucket，只需要增加这个bucket内count的数字，而不用担心buckets[]所占用空间大小的剧增。姑且把它称为Window Buckets方法。
考虑Concurrent, Distributed
Solution
Straightforward Queue Inplementation
class HitCounter {
ArrayDeque trac;
/** Initialize your data structure here. */
private final int FIVE_MINUTES = 300;
public HitCounter() {
trac = new ArrayDeque<Integer>();
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
trac.addLast(timestamp);
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
while(trac.size() > 0 && ( int) trac.getFirst() + FIVE_MINUTES <= timestamp) {
trac.removeFirst();
} return trac.size();
}
}
Queue
class HitCounter {
Queue<Integer> q;
/** Initialize your data structure here. */
private final int FIVE_MINUTES = 300;
public HitCounter() {
q = new LinkedList<Integer>();
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity).
*/
public void hit(int timestamp)
{
q.offer(timestamp);
}
// Time Complexity : O(1)
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds
granularity).
*/
public int getHits(int timestamp)
{
while (!q.isEmpty() && (timestamp - q.peek() >= FIVE_MINUTES)) {
q.poll();
}
return q.size();
}
// Time Complexity : O(n)
}
/**
* Your HitCounter object will be instantiated and called as such:
* HitCounter obj = new HitCounter();
* obj.hit(timestamp);
* int param_2 = obj.getHits(timestamp);
*/
Time: hit() - O(1), getHits() - O(M) where the M is the window size
Space: O(M) - the M is window size
class HitCounter {
int[] hits;
int[] times;
int WINDOW_SIZE = 300;
/** Initialize your data structure here. */
public HitCounter() {
hits = new int[WINDOW_SIZE];
times = new int[WINDOW_SIZE];
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
int idx = timestamp % WINDOW_SIZE;
if (times[idx] != timestamp) {
times[idx] = timestamp;
hits[idx] = 1;
} else {
hits[idx] += 1;
}
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
int count = 0;
for (int i = 0; i < WINDOW_SIZE; i++) {
if (timestamp - times[i] < 300) {
count += hits[i];
}
}
return count;
}
}
/**
* Your HitCounter object will be instantiated and called as such:
* HitCounter obj = new HitCounter();
* obj.hit(timestamp);
* int param_2 = obj.getHits(timestamp);
*/