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 <= 1051 <= arrival.length, load.length <= 105arrival.length == load.length1 <= arrival[i], load[i] <= 109arrival is strictly increasing.
Solutions#
Solution 1#
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| 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]
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| 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;
}
}
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| 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;
}
};
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| 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 }
|
