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57 lines
3.4 KiB
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57 lines
3.4 KiB
HTML
<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>
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<ul>
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<li>The <code>i<sup>th</sup></code> (0-indexed) request arrives.</li>
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<li>If all servers are busy, the request is dropped (not handled at all).</li>
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<li>If the <code>(i % k)<sup>th</sup></code> server is available, assign the request to that server.</li>
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<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>
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</ul>
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<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>
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<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>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/09/08/load-1.png" style="width: 389px; height: 221px;" />
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<pre>
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<strong>Input:</strong> k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]
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<strong>Output:</strong> [1]
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<strong>Explanation:</strong>
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All of the servers start out available.
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The first 3 requests are handled by the first 3 servers in order.
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Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
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Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
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Servers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
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<strong>Output:</strong> [0]
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<strong>Explanation:</strong>
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The first 3 requests are handled by first 3 servers.
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Request 3 comes in. It is handled by server 0 since the server is available.
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Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> k = 3, arrival = [1,2,3], load = [10,12,11]
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<strong>Output:</strong> [0,1,2]
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<strong>Explanation:</strong> Each server handles a single request, so they are all considered the busiest.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= k <= 10<sup>5</sup></code></li>
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<li><code>1 <= arrival.length, load.length <= 10<sup>5</sup></code></li>
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<li><code>arrival.length == load.length</code></li>
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<li><code>1 <= arrival[i], load[i] <= 10<sup>9</sup></code></li>
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<li><code>arrival</code> is <strong>strictly increasing</strong>.</li>
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</ul>
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