# 1606. Find Servers That Handled Most Number of Requests

<https://leetcode.com/problems/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:**

![](https://assets.leetcode.com/uploads/2020/09/08/load-1.png)

```
**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.
```

**Example 4:**

```
**Input:** k = 3, arrival = [1,2,3,4,8,9,10], load = [5,2,10,3,1,2,2]
**Output:** [1]
```

**Example 5:**

```
**Input:** k = 1, arrival = [1], load = [1]
**Output:** [0]
```

**Constraints:**

* `1 <= k <= 105`
* `1 <= arrival.length, load.length <= 105`
* `arrival.length == load.length`
* `1 <= arrival[i], load[i] <= 109`
* `arrival` is **strictly increasing**.

## ac

```java
```