Solution Explanation

This problem involves assigning incoming requests to servers, considering server availability and request load. The goal is to identify the servers that handled the most requests. The solution uses a combination of data structures to efficiently manage server availability and request processing.

Approach:

1. Data Structures:

   • free: A sorted set (or equivalent) to store the IDs of available servers. A sorted set provides efficient insertion, deletion, and searching for the next available server.
   • busy: A min-heap (priority queue) to store the currently busy servers. Each element in the heap contains the server's ID and its release time. The min-heap ensures we can quickly identify the next available server.
   • cnt: An array to count the number of requests handled by each server.

2. Request Processing:

   • Iterate through each request (arrival time, load).
   • Check for Available Servers: Remove any servers from the busy heap that have finished processing their requests (release time <= current arrival time). Add these servers back to the free set.
   • Assign Request:
     • If there are available servers: Assign the request to the next available server using the following approach:
       • Ideally, we'd assign it to i % k, but if that server is busy, we use the sorted set free to find the next available server in a wrapped-around manner (starting from i % k).
     • If no servers are available, drop the request (this is handled automatically since free will be empty).
   • Update States: Add the currently assigned server to the busy heap with its release time, remove the server from free, and increment its request count in cnt.

3. Find Busiest Servers: After processing all requests, find the maximum number of requests handled among all servers (mx). Then, return the list of server IDs that have handled mx requests.

Time Complexity:

The time complexity is dominated by the iteration through the requests. Each operation on the free sorted set and the busy min-heap (insertion, deletion, searching) takes O(log k) time, where k is the number of servers. Therefore, the overall time complexity is O(N log k), where N is the number of requests.

Space Complexity:

The space complexity is O(k) to store free, busy, and cnt. This is because the number of servers is fixed and independent of the number of requests.

Code Explanation (Python):

The Python code utilizes the SortedList from the sortedcontainers library (a more efficient sorted set implementation than Python's built-in set). The code follows the algorithm described above:

from sortedcontainers import SortedList
import heapq
 
class Solution:
    def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
        free = SortedList(range(k))
        busy = [] # Use a list as a min-heap
        cnt = [0] * k
        for i, (start, t) in enumerate(zip(arrival, load)):
            while busy and busy[0][0] <= start:
                free.add(busy[0][1])
                heapq.heappop(busy) # Correct heappop usage
            if not free:
                continue
            j = free.bisect_left(i % k)
            if j == len(free):
                j = 0
            server = free[j]
            cnt[server] += 1
            heapq.heappush(busy, (start + t, server)) # Correct heappush usage
            free.remove(server)
 
        mx = max(cnt)
        return [i for i, v in enumerate(cnt) if v == mx]

The other languages (Java, C++, Go) implement similar logic, using their respective priority queue and sorted set implementations. Note that Go's solution uses a custom heap and redblacktree for efficient priority queue and sorted set functionality. Java uses the standard PriorityQueue and TreeSet. C++ leverages priority_queue and set.