Question

1606. Find Servers That Handled Most Number of Requests

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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

from sortedcontainers import SortedList


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]

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;
  }
}

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;
  }
};

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 }