Class Solution

java.lang.Object
g0501_0600.s0502_ipo.Solution

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

    • Solution

      public Solution()
  • Method Details

    • findMaximizedCapital

      public int findMaximizedCapital(int k, int w, int[] profits, int[] capital)