Skip to content

1606. Find Servers That Handled Most Number of Requests

Description

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.

Solutions

Solution 1

Thinking

Both \(k\) and the number of requests can be \(10^5\), so scanning all servers per request is too slow. We need the idle server with id at least \(i \bmod k\), wrapping to the smallest id if none exists.

Busy servers should rejoin the idle set when they finish, so a heap stores \((\textit{end},\textit{server})\) and an ordered set stores free ids.

On each arrival, release finished jobs, then \(\texttt{bisect\_left}(i \bmod k)\) in \(\textit{free}\). Array \(\textit{cnt}\) records load; the answer is every id that attains the maximum.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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]

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
class Solution {
    public List<Integer> busiestServers(int k, int[] arrival, int[] load) {
        int[] cnt = new int[k];
        PriorityQueue<int[]> busy = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        TreeSet<Integer> free = new TreeSet<>();
        for (int i = 0; i < k; ++i) {
            free.add(i);
        }
        for (int i = 0; i < arrival.length; ++i) {
            int start = arrival[i];
            int end = start + load[i];
            while (!busy.isEmpty() && busy.peek()[0] <= start) {
                free.add(busy.poll()[1]);
            }
            if (free.isEmpty()) {
                continue;
            }
            Integer server = free.ceiling(i % k);
            if (server == null) {
                server = free.first();
            }
            ++cnt[server];
            busy.offer(new int[] {end, server});
            free.remove(server);
        }
        int mx = 0;
        for (int v : cnt) {
            mx = Math.max(mx, v);
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < k; ++i) {
            if (cnt[i] == mx) {
                ans.add(i);
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
class Solution {
public:
    vector<int> busiestServers(int k, vector<int>& arrival, vector<int>& load) {
        set<int> free;
        for (int i = 0; i < k; ++i) free.insert(i);
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> busy;
        vector<int> cnt(k);
        for (int i = 0; i < arrival.size(); ++i) {
            int start = arrival[i], end = start + load[i];
            while (!busy.empty() && busy.top().first <= start) {
                free.insert(busy.top().second);
                busy.pop();
            }
            if (free.empty()) continue;
            auto p = free.lower_bound(i % k);
            if (p == free.end()) p = free.begin();
            int server = *p;
            ++cnt[server];
            busy.emplace(end, server);
            free.erase(server);
        }
        int mx = *max_element(cnt.begin(), cnt.end());
        vector<int> ans;
        for (int i = 0; i < k; ++i)
            if (cnt[i] == mx)
                ans.push_back(i);
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
func busiestServers(k int, arrival, load []int) (ans []int) {
    free := redblacktree.NewWithIntComparator()
    for i := 0; i < k; i++ {
        free.Put(i, nil)
    }
    busy := hp{}
    cnt := make([]int, k)
    for i, t := range arrival {
        for len(busy) > 0 && busy[0].end <= t {
            free.Put(busy[0].server, nil)
            heap.Pop(&busy)
        }
        if free.Size() == 0 {
            continue
        }
        p, _ := free.Ceiling(i % k)
        if p == nil {
            p = free.Left()
        }
        server := p.Key.(int)
        cnt[server]++
        heap.Push(&busy, pair{t + load[i], server})
        free.Remove(server)
    }
    mx := slices.Max(cnt)
    for i, v := range cnt {
        if v == mx {
            ans = append(ans, i)
        }
    }
    return
}

type pair struct{ end, server int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

Comments