#1606

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Find Servers That Handled Most Number of Requests

Given k servers and a series of requests, find the busiest server(s) using greedy strategies and efficient server tracking.

array greedy heap priority queue ordered set

Problem Statement

You are given k servers that can handle multiple requests but can process only one request at a time. Requests arrive at strictly increasing times, and each request has a processing time. Your task is to find the server(s) that handled the most requests.

For each request, you must assign it to the next available server. If no servers are free, the request is dropped. The goal is to return the IDs of the server(s) that handled the most requests. Servers should be tracked efficiently to determine availability quickly.

Examples

Example 1

Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]

Output: [1]

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]

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]

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.

Solution Approach

Greedy Assignment with Heap

We use a greedy strategy, processing requests as they arrive and assigning each to the next available server. A heap (priority queue) keeps track of available servers, allowing quick access to the next available one.

Efficient Server Availability Tracking

To ensure optimal assignment, we track the servers' availability using a sorted structure, such as an ordered set or a heap. This ensures that the search for the next available server is efficient, even with a large number of requests.

Counting Requests Handled

Once requests are assigned, we count the number of requests handled by each server. The server(s) that handled the most requests are returned as the answer.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used to track server availability. Using a heap, the complexity for each request is O(log k), leading to a total time complexity of O(n log k), where n is the number of requests. The space complexity is O(k) for tracking server availability and request handling counts.

What Interviewers Usually Probe

Evaluating the use of heaps or ordered sets to manage server availability indicates a strong understanding of efficient data structures.
Attention to handling edge cases such as all servers being busy or requests being dropped tests the candidate's problem-solving thoroughness.
Efficiently managing a large number of requests signals competency in optimizing both space and time complexity.

Common Pitfalls or Variants

Common pitfalls

Incorrectly handling requests when no server is available, leading to errors in dropped request counting.
Failure to maintain proper availability of servers, potentially causing incorrect assignment of requests.
Not considering edge cases such as when all servers handle the same number of requests, leading to wrong results.

Follow-up variants

Change the server handling capacity, such as allowing multiple requests to be processed simultaneously on each server.
Introduce varying server speeds, where each server has a different handling time for requests, requiring dynamic load balancing.
Implement a scenario where requests can be delayed or split, complicating the availability of servers and request completion.

FAQ

How does greedy choice work in this problem?

Greedy choice ensures that each incoming request is assigned to the next available server, aiming to process the most requests with the available servers.

What is the best way to track server availability?

Using a heap or an ordered set is ideal for tracking server availability efficiently, as it allows quick access to the next available server.

How can I handle the case when no server is available?

When no server is available, the request should be dropped, and you should ensure this case is properly handled in the logic.

How do I ensure the solution scales well with a large number of requests?

To scale well, optimize the server tracking and request assignment using efficient data structures, such as heaps, which provide O(log k) complexity for server availability updates.

What happens if multiple servers handle the same number of requests?

In this case, any of the servers that handled the maximum number of requests can be returned, as the problem allows multiple valid answers.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
        free = SortedList(range(k))
        busy = []
        cnt = [0] * k
        for i, (start, t) in enumerate(zip(arrival, load)):
            while busy and busy[0][0] <= start:
                free.add(busy[0][1])
                heappop(busy)
            if not free:
                continue
            j = free.bisect_left(i % k)
            if j == len(free):
                j = 0
            server = free[j]
            cnt[server] += 1
            heappush(busy, (start + t, server))
            free.remove(server)

        mx = max(cnt)
        return [i for i, v in enumerate(cnt) if v == mx]

Continue Practicing

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