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43 lines
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43 lines
2.3 KiB
HTML
<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>
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<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>
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<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>
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<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>
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<p>The answer is guaranteed to fit in a 32-bit signed integer.</p>
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<p> </p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Since your initial capital is 0, you can only start the project indexed 0.
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After finishing it you will obtain profit 1 and your capital becomes 1.
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With capital 1, you can either start the project indexed 1 or the project indexed 2.
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Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.
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Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.
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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, w = 0, profits = [1,2,3], capital = [0,1,2]
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<strong>Output:</strong> 6
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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>0 <= w <= 10<sup>9</sup></code></li>
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<li><code>n == profits.length</code></li>
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<li><code>n == capital.length</code></li>
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<li><code>1 <= n <= 10<sup>5</sup></code></li>
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<li><code>0 <= profits[i] <= 10<sup>4</sup></code></li>
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<li><code>0 <= capital[i] <= 10<sup>9</sup></code></li>
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</ul>
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