1606 Find Servers That Handled Most Number of Requests

You have k servers numbered from 0 to k-1 that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time. The requests are assigned to servers according to a specific algorithm:

• The ith (0-indexed) request arrives.
• If all servers are busy, the request is dropped (not handled at all).
• If the (i % k)th server is available, assign the request to that server.
• Otherwise, assign the request to the next available server (wrapping around the list of servers and starting from 0 if necessary). For example, if the ith server is busy, try to assign the request to the (i+1)th server, then the (i+2)th server, and so on.

You are given a strictly increasing array arrival of positive integers, where arrival[i] represents the arrival time of the ith request, and another array load, where load[i] represents the load of the ith request (the time it takes to complete). Your goal is to find the busiest server(s). A server is considered busiest if it handled the most number of requests successfully among all the servers.

Return a list containing the IDs (0-indexed) of the busiest server(s). You may return the IDs in any order.

Example 1:

Dropped                   [R4--L=3----]
Server 2          [R2--L=3----]
Server 1     [R1--L=2-][R3--L=3---]
Server 0  [R0--L=5------------]
          1   2   3   4   5   6   7   8

Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
Output: [1] 
Explanation: 
All of the servers start out available.
The first 3 requests are handled by the first 3 servers in order.
Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
Servers 0 and 2 handled one request each, while server 1 handled two requests.
Hence server 1 is the busiest server.

Example 2:

Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
Output: [0]
Explanation: 
The first 3 requests are handled by first 3 servers.
Request 3 comes in. It is handled by server 0 since the server is available.
Server 0 handled two requests, while servers 1 and 2 handled one request each.
Hence server 0 is the busiest server.

Example 3:

Input: k = 3, arrival = [1,2,3], load = [10,12,11]
Output: [0,1,2]
Explanation: Each server handles a single request,
so they are all considered the busiest.

Constraints:

• 1 <= k <= 105
• 1 <= arrival.length, load.length <= 105
• arrival.length == load.length
• 1 <= arrival[i], load[i] <= 109
• arrival is strictly increasing.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

class Solution:
    def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
        count = [0] * k
        busy = []  # min heap (end_time, server_id)
        free = list(range(k))  # min heap of free servers
        
        for i, (a, l) in enumerate(zip(arrival, load)):
            while busy and busy[0][0] <= a:
                # release finished servers
                _, server_id = heapq.heappop(busy)
                # make sure idx >= i
                idx = i + (server_id - i) % k
                heapq.heappush(free, idx)

            if not free:
                continue

            busy_id = heapq.heappop(free) % k
            heapq.heappush(busy, (a + l, busy_id))
            count[busy_id] += 1

        max_count = max(count)
        return [i for i, c in enumerate(count) if c == max_count]