mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
186 lines
24 KiB
JSON
186 lines
24 KiB
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"content": "<p>Suppose LeetCode will start its <strong>IPO</strong> soon. In order to sell a good price of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the <strong>IPO</strong>. Since it has limited resources, it can only finish at most <code>k</code> distinct projects before the <strong>IPO</strong>. Help LeetCode design the best way to maximize its total capital after finishing at most <code>k</code> distinct projects.</p>\n\n<p>You are given <code>n</code> projects where the <code>i<sup>th</sup></code> project has a pure profit <code>profits[i]</code> and a minimum capital of <code>capital[i]</code> is needed to start it.</p>\n\n<p>Initially, you have <code>w</code> capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.</p>\n\n<p>Pick a list of <strong>at most</strong> <code>k</code> distinct projects from given projects to <strong>maximize your final capital</strong>, and return <em>the final maximized capital</em>.</p>\n\n<p>The answer is guaranteed to fit in a 32-bit signed integer.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> Since your initial capital is 0, you can only start the project indexed 0.\nAfter finishing it you will obtain profit 1 and your capital becomes 1.\nWith capital 1, you can either start the project indexed 1 or the project indexed 2.\nSince you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.\nTherefore, output the final maximized capital, which is 0 + 1 + 3 = 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]\n<strong>Output:</strong> 6\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= k <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= w <= 10<sup>9</sup></code></li>\n\t<li><code>n == profits.length</code></li>\n\t<li><code>n == capital.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= profits[i] <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= capital[i] <= 10<sup>9</sup></code></li>\n</ul>\n",
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"translatedContent": "<p>假设 力扣(LeetCode)即将开始 <strong>IPO</strong> 。为了以更高的价格将股票卖给风险投资公司,力扣 希望在 IPO 之前开展一些项目以增加其资本。 由于资源有限,它只能在 IPO 之前完成最多 <code>k</code> 个不同的项目。帮助 力扣 设计完成最多 <code>k</code> 个不同项目后得到最大总资本的方式。</p>\n\n<p>给你 <code>n</code> 个项目。对于每个项目 <code>i</code><strong> </strong>,它都有一个纯利润 <code>profits[i]</code> ,和启动该项目需要的最小资本 <code>capital[i]</code> 。</p>\n\n<p>最初,你的资本为 <code>w</code> 。当你完成一个项目时,你将获得纯利润,且利润将被添加到你的总资本中。</p>\n\n<p>总而言之,从给定项目中选择 <strong>最多</strong> <code>k</code> 个不同项目的列表,以 <strong>最大化最终资本</strong> ,并输出最终可获得的最多资本。</p>\n\n<p>答案保证在 32 位有符号整数范围内。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]\n<strong>输出:</strong>4\n<strong>解释:\n</strong>由于你的初始资本为 0,你仅可以从 0 号项目开始。\n在完成后,你将获得 1 的利润,你的总资本将变为 1。\n此时你可以选择开始 1 号或 2 号项目。\n由于你最多可以选择两个项目,所以你需要完成 2 号项目以获得最大的资本。\n因此,输出最后最大化的资本,为 0 + 1 + 3 = 4。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]\n<strong>输出:</strong>6\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= k <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= w <= 10<sup>9</sup></code></li>\n\t<li><code>n == profits.length</code></li>\n\t<li><code>n == capital.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= profits[i] <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= capital[i] <= 10<sup>9</sup></code></li>\n</ul>\n",
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