# 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]``````