leetcode-problemset/leetcode-cn/originData/maximum-number-of-robots-within-budget.json

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            "questionId": "2449",
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            "categoryTitle": "Algorithms",
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            "title": "Maximum Number of Robots Within Budget",
            "titleSlug": "maximum-number-of-robots-within-budget",
            "content": "<p>You have <code>n</code> robots. You are given two <strong>0-indexed</strong> integer arrays, <code>chargeTimes</code> and <code>runningCosts</code>, both of length <code>n</code>. The <code>i<sup>th</sup></code> robot costs <code>chargeTimes[i]</code> units to charge and costs <code>runningCosts[i]</code> units to run. You are also given an integer <code>budget</code>.</p>\n\n<p>The <strong>total cost</strong> of running <code>k</code> chosen robots is equal to <code>max(chargeTimes) + k * sum(runningCosts)</code>, where <code>max(chargeTimes)</code> is the largest charge cost among the <code>k</code> robots and <code>sum(runningCosts)</code> is the sum of running costs among the <code>k</code> robots.</p>\n\n<p>Return<em> the <strong>maximum</strong> number of <strong>consecutive</strong> robots you can run such that the total cost <strong>does not</strong> exceed </em><code>budget</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> \nIt is possible to run all individual and consecutive pairs of robots within budget.\nTo obtain answer 3, consider the first 3 robots. The total cost will be max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 which is less than 25.\nIt can be shown that it is not possible to run more than 3 consecutive robots within budget, so we return 3.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> No robot can be run that does not exceed the budget, so we return 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>chargeTimes.length == runningCosts.length == n</code></li>\n\t<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= chargeTimes[i], runningCosts[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= budget &lt;= 10<sup>15</sup></code></li>\n</ul>\n",
            "translatedTitle": "预算内的最多机器人数目",
            "translatedContent": "<p>你有&nbsp;<code>n</code>&nbsp;个机器人，给你两个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>chargeTimes</code> 和&nbsp;<code>runningCosts</code>&nbsp;，两者长度都为&nbsp;<code>n</code>&nbsp;。第&nbsp;<code>i</code>&nbsp;个机器人充电时间为&nbsp;<code>chargeTimes[i]</code>&nbsp;单位时间，花费&nbsp;<code>runningCosts[i]</code>&nbsp;单位时间运行。再给你一个整数&nbsp;<code>budget</code>&nbsp;。</p>\n\n<p>运行&nbsp;<code>k</code>&nbsp;个机器人 <strong>总开销</strong>&nbsp;是&nbsp;<code>max(chargeTimes) + k * sum(runningCosts)</code>&nbsp;，其中&nbsp;<code>max(chargeTimes)</code>&nbsp;是这&nbsp;<code>k</code>&nbsp;个机器人中最大充电时间，<code>sum(runningCosts)</code>&nbsp;是这 <code>k</code>&nbsp;个机器人的运行时间之和。</p>\n\n<p>请你返回在 <strong>不超过</strong>&nbsp;<code>budget</code>&nbsp;的前提下，你 <strong>最多</strong>&nbsp;可以 <strong>连续</strong>&nbsp;运行的机器人数目为多少。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25\n<b>输出：</b>3\n<b>解释：</b>\n可以在 budget 以内运行所有单个机器人或者连续运行 2 个机器人。\n选择前 3 个机器人，可以得到答案最大值 3 。总开销是 max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 ，小于 25 。\n可以看出无法在 budget 以内连续运行超过 3 个机器人，所以我们返回 3 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19\n<b>输出：</b>0\n<b>解释：</b>即使运行任何一个单个机器人，还是会超出 budget，所以我们返回 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>chargeTimes.length == runningCosts.length == n</code></li>\n\t<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= chargeTimes[i], runningCosts[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= budget &lt;= 10<sup>15</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {\n        \n    }\n};",
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                "Use binary search to convert the problem into checking if we can find a specific number of consecutive robots within the budget.",
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