leetcode-problemset/leetcode/originData/minimum-processing-time.json

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            "title": "Minimum Processing Time",
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            "content": "<p>You have <code>n</code> processors each having <code>4</code> cores and <code>n * 4</code> tasks that need to be executed such that each core should perform only <strong>one</strong> task.</p>\n\n<p>Given a <strong>0-indexed</strong> integer array <code>processorTime</code> representing the time at which each processor becomes available for the first time and a <strong>0-indexed </strong>integer array <code>tasks</code> representing the time it takes to execute each task, return <em>the <strong>minimum</strong> time when all of the tasks have been executed by the processors.</em></p>\n\n<p><strong>Note: </strong>Each core executes the task independently of the others.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> processorTime = [8,10], tasks = [2,2,3,1,8,7,4,5]\n<strong>Output:</strong> 16\n<strong>Explanation:</strong> \nIt&#39;s optimal to assign the tasks at indexes 4, 5, 6, 7 to the first processor which becomes available at time = 8, and the tasks at indexes 0, 1, 2, 3 to the second processor which becomes available at time = 10. \nTime taken by the first processor to finish execution of all tasks = max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16.\nTime taken by the second processor to finish execution of all tasks = max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13.\nHence, it can be shown that the minimum time taken to execute all the tasks is 16.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> processorTime = [10,20], tasks = [2,3,1,2,5,8,4,3]\n<strong>Output:</strong> 23\n<strong>Explanation:</strong> \nIt&#39;s optimal to assign the tasks at indexes 1, 4, 5, 6 to the first processor which becomes available at time = 10, and the tasks at indexes 0, 2, 3, 7 to the second processor which becomes available at time = 20.\nTime taken by the first processor to finish execution of all tasks = max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18.\nTime taken by the second processor to finish execution of all tasks = max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23.\nHence, it can be shown that the minimum time taken to execute all the tasks is 23.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == processorTime.length &lt;= 25000</code></li>\n\t<li><code>1 &lt;= tasks.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= processorTime[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= tasks[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>tasks.length == 4 * n</code></li>\n</ul>\n",
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                    "code": "class Solution(object):\n    def minProcessingTime(self, processorTime, tasks):\n        \"\"\"\n        :type processorTime: List[int]\n        :type tasks: List[int]\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public int MinProcessingTime(IList<int> processorTime, IList<int> tasks) {\n        \n    }\n}",
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                "It’s optimal to make the processor with earlier process time run 4 longer tasks.****",
                "The largest <code>processTime[i] + tasks[j]</code> (when matched) is the answer."
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