k servers numbered from
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:
ith(0-indexed) request arrives.
- If all servers are busy, the request is dropped (not handled at all).
- If the
(i % k)thserver 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
ithserver is busy, try to assign the request to the
(i+1)thserver, then the
(i+2)thserver, 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[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.
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:  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.
Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2] Output:  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.
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.
1 <= k <= 105
1 <= arrival.length, load.length <= 105
arrival.length == load.length
1 <= arrival[i], load[i] <= 109
arrivalis strictly increasing.