-
Notifications
You must be signed in to change notification settings - Fork 25
/
Copy path1606-find-servers-that-handled-most-number-of-requests.js
101 lines (88 loc) · 3.14 KB
/
1606-find-servers-that-handled-most-number-of-requests.js
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
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
/**
* 1606. Find Servers That Handled Most Number of Requests
* https://leetcode.com/problems/find-servers-that-handled-most-number-of-requests/
* Difficulty: 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.
*/
/**
* @param {number} k
* @param {number[]} arrival
* @param {number[]} load
* @return {number[]}
*/
var busiestServers = function(k, arrival, load) {
const requests = new Array(k).fill(0);
const busy = new PriorityQueue((a, b) => a[0] - b[0]);
const right = Array.from({length: k}, (_, i) => i);
let left = [];
for (let i = 0; i < arrival.length; i++) {
const time = arrival[i];
const target = i % k;
while (!busy.isEmpty() && busy.front()[0] <= time) {
const serverId = busy.dequeue()[1];
(serverId >= target) ? binaryInsert(right, serverId) : binaryInsert(left, serverId);
}
let assigned = -1;
if (right.length > 0) {
const index = binarySearch(right, target);
if (index < right.length) {
assigned = right[index];
right.splice(index, 1);
}
}
if (assigned === -1 && left.length > 0) {
assigned = left[0];
left.shift();
}
if (assigned !== -1) {
requests[assigned]++;
busy.enqueue([time + load[i], assigned]);
}
if ((i + 1) % k === 0) {
right.push(...left);
right.sort((a, b) => a - b);
left = [];
}
}
const maxRequests = Math.max(...requests);
return requests.reduce((result, count, index) => {
if (count === maxRequests) result.push(index);
return result;
}, []);
};
function binarySearch(arr, target) {
let left = 0;
let right = arr.length - 1;
let result = arr.length;
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (arr[mid] >= target) {
result = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
function binaryInsert(arr, val) {
arr.splice(binarySearch(arr, val), 0, val);
}