Introduction

In this blog post, we will explore how to solve the LeetCode problem "1606" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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: 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.

Explanation

Here's the solution to the problem:

• Key Idea: Use a priority queue (min-heap) to track available servers and their finish times. This allows efficient retrieval of the next available server. Also, maintain a count of requests handled by each server to determine the busiest server(s).

• Simulation: Iterate through the requests, and for each request, check if any servers have become available since the last request. If so, add them back to the available servers' heap. Find the next available server using modular arithmetic and the available servers' heap.

• Busiest Server Calculation: Keep track of the number of requests each server handles. After processing all requests, identify and return the servers that handled the maximum number of requests.

• Complexity:

  • Runtime Complexity: O(n log k), where n is the number of requests and k is the number of servers.
  • Storage Complexity: O(k)

Code

    import heapq

def busiestServers(k: int, arrival: list[int], load: list[int]) -> list[int]:
    """
    Finds the busiest server(s) that handled the most number of requests.

    Args:
        k: The number of servers.
        arrival: A list of arrival times of requests.
        load: A list of loads (processing times) of requests.

    Returns:
        A list of the IDs of the busiest server(s).
    """

    available_servers = list(range(k))  # List of available server indices
    heapq.heapify(available_servers) # transform list into heap, in-place, in linear time
    busy_servers = []  # List of tuples: (finish_time, server_index)
    requests_handled = [0] * k  # Count of requests handled by each server

    for i in range(len(arrival)):
        while busy_servers and busy_servers[0][0] <= arrival[i]:
            _, server_index = heapq.heappop(busy_servers)
            heapq.heappush(available_servers, server_index)

        if not available_servers:
            continue  # Drop the request if no servers are available

        start_server_index = i % k
        found_server = False
        temp_available_servers = [] # use this array to temporarily hold available servers so that servers can be sorted

        while available_servers: # if available_servers is not empty
          server_index = heapq.heappop(available_servers)

          if server_index >= start_server_index:
            heapq.heappush(busy_servers, (arrival[i] + load[i], server_index))
            requests_handled[server_index] += 1
            found_server = True
            break

          else:
            temp_available_servers.append(server_index)

        if not found_server:
          while temp_available_servers:
              server_index = temp_available_servers.pop()
              heapq.heappush(available_servers, server_index)

          while available_servers: # if available_servers is not empty
            server_index = heapq.heappop(available_servers)
            heapq.heappush(busy_servers, (arrival[i] + load[i], server_index))
            requests_handled[server_index] += 1
            found_server = True
            break

        while temp_available_servers: # push back other available servers for the next loop
          server_index = temp_available_servers.pop()
          heapq.heappush(available_servers, server_index)

    max_requests = max(requests_handled)
    busiest_servers = [i for i, count in enumerate(requests_handled) if count == max_requests]

    return busiest_servers