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            "title": "Find Servers That Handled Most Number of Requests",
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            "content": "<p>You have <code>k</code> servers numbered from <code>0</code> to <code>k-1</code> that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but <strong>cannot handle more than one request at a time</strong>. The requests are assigned to servers according to a specific algorithm:</p>\n\n<ul>\n\t<li>The <code>i<sup>th</sup></code> (0-indexed) request arrives.</li>\n\t<li>If all servers are busy, the request is dropped (not handled at all).</li>\n\t<li>If the <code>(i % k)<sup>th</sup></code> server is available, assign the request to that server.</li>\n\t<li>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 <code>i<sup>th</sup></code> server is busy, try to assign the request to the <code>(i+1)<sup>th</sup></code> server, then the <code>(i+2)<sup>th</sup></code> server, and so on.</li>\n</ul>\n\n<p>You are given a <strong>strictly increasing</strong> array <code>arrival</code> of positive integers, where <code>arrival[i]</code> represents the arrival time of the <code>i<sup>th</sup></code> request, and another array <code>load</code>, where <code>load[i]</code> represents the load of the <code>i<sup>th</sup></code> request (the time it takes to complete). Your goal is to find the <strong>busiest server(s)</strong>. A server is considered <strong>busiest</strong> if it handled the most number of requests successfully among all the servers.</p>\n\n<p>Return <em>a list containing the IDs (0-indexed) of the <strong>busiest server(s)</strong></em>. You may return the IDs in any order.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/09/08/load-1.png\" style=\"width: 389px; height: 221px;\" />\n<pre>\n<strong>Input:</strong> k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] \n<strong>Output:</strong> [1] \n<strong>Explanation:</strong> \nAll of the servers start out available.\nThe first 3 requests are handled by the first 3 servers in order.\nRequest 3 comes in. Server 0 is busy, so it&#39;s assigned to the next available server, which is 1.\nRequest 4 comes in. It cannot be handled since all servers are busy, so it is dropped.\nServers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 3, arrival = [1,2,3,4], load = [1,2,1,2]\n<strong>Output:</strong> [0]\n<strong>Explanation:</strong> \nThe first 3 requests are handled by first 3 servers.\nRequest 3 comes in. It is handled by server 0 since the server is available.\nServer 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 3, arrival = [1,2,3], load = [10,12,11]\n<strong>Output:</strong> [0,1,2]\n<strong>Explanation:</strong> Each server handles a single request, so they are all considered the busiest.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= arrival.length, load.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>arrival.length == load.length</code></li>\n\t<li><code>1 &lt;= arrival[i], load[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>arrival</code> is <strong>strictly increasing</strong>.</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> busiestServers(int k, vector<int>& arrival, vector<int>& load) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def busiestServers(self, k, arrival, load):\n        \"\"\"\n        :type k: int\n        :type arrival: List[int]\n        :type load: List[int]\n        :rtype: List[int]\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public IList<int> BusiestServers(int k, int[] arrival, int[] load) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number} k\n * @param {number[]} arrival\n * @param {number[]} load\n * @return {number[]}\n */\nvar busiestServers = function(k, arrival, load) {\n    \n};",
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                    "code": "# @param {Integer} k\n# @param {Integer[]} arrival\n# @param {Integer[]} load\n# @return {Integer[]}\ndef busiest_servers(k, arrival, load)\n    \nend",
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                    "code": "impl Solution {\n    pub fn busiest_servers(k: i32, arrival: Vec<i32>, load: Vec<i32>) -> Vec<i32> {\n        \n    }\n}",
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                "To speed up the next available server search, keep track of the available servers in a sorted structure such as an ordered set.",
                "To determine if a server is available, keep track of the end times for each task in a heap and add the server to the available set once the soonest task ending time is less than or equal to the next task to add."
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