LeetCode in Kotlin

1606. Find Servers That Handled Most Number of Requests

Hard

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.

Solution

import java.util.PriorityQueue
import java.util.TreeSet

class Solution {
    internal class Server(val id: Int, val busyTime: Int)

    fun busiestServers(k: Int, arrival: IntArray, load: IntArray): List<Int> {
        val available = TreeSet<Int>()
        val busy = PriorityQueue({ a: Server, b: Server -> a.busyTime.compareTo(b.busyTime) })
        val requestCount = IntArray(k)
        val n = arrival.size
        for (id in 0 until k) {
            available.add(id)
        }
        for (i in 0 until n) {
            val defaultServer = i % k
            while (busy.isNotEmpty() && busy.peek().busyTime <= arrival[i]) {
                val top = busy.poll()
                available.add(top.id)
            }
            if (available.isEmpty()) {
                continue
            }
            var nextServer = available.ceiling(defaultServer)
            nextServer = nextServer ?: available.ceiling(0)
            val requestEnd = arrival[i] + load[i]
            available.remove(nextServer)
            busy.add(Server(nextServer, requestEnd))
            requestCount[nextServer]++
        }
        var maxRequests = Int.MIN_VALUE
        val busiestServers: MutableList<Int> = ArrayList()
        for (id in 0 until k) {
            maxRequests = Math.max(maxRequests, requestCount[id])
        }
        for (id in 0 until k) {
            if (requestCount[id] == maxRequests) {
                busiestServers.add(id)
            }
        }
        return busiestServers
    }
}

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