This is the daily question of 2021-9-8 on LeetCode: 「502. IPO」

# 1. Title Description

Suppose LeetCode is about to start IPO. In order to sell shares to venture capital companies at a higher price, Li Kou hopes to carry out some projects to increase its capital before IPO. Due to limited resources, it can only complete up to k different projects before IPO. Help force buckle design the way to get the maximum total capital after completing up to k different projects.

Here are n items for you. For each project I, it has a net profit profits[i], and the minimum capital[i] required to start the project.

Initially, your capital was w. When you complete a project, you will get a net profit, and the profit will be added to your total capital.

In summary, select a list of up to k different projects from a given project to maximize the final capital and output the maximum capital available.

The answer is guaranteed to be in the range of 32-bit signed integers.

Example 1:

Input: k = 2, w = 0, profits = [1,2,3], capital = [0,1,1] Output: 4 Explanation: Since your initial capital is 0, you can only start from project 0. After completion, you will get a profit of 1 and your total capital will become 1. At this point, you can choose to start project 1 or 2. Since you can select up to two projects, you need to complete project 2 to get the maximum capital. Therefore, the output of the last maximized capital is 0 + 1 + 3 = 4.

Example 2:

Input: k = 3, w = 0, profits = [1,2,3], capital = [0,1,2] Output: 6

# 2. Answer

Greedy thought: the net profit of each investment project should be the largest

- First obtain the item array arr and combine the capital and profit by item number
- Sort the projects according to the required capital from small to large (if from large to small, it may be impossible to invest at the beginning)
- Insert the profits from all qualified projects into the maximum heap
- If the heap is not empty, take out the top of the heap, which is the maximum net profit and update your own capital w
- If the heap is empty, exit the loop directly because there are no items that meet the conditions to enter the heap
- Repeat steps 3 to 5 for a total of k times
- Finally, return the final capital w

// Default maximum heap const defaultCmp = (x, y) => x > y; // Exchange element const swap = (arr, i, j) => ([arr[i], arr[j]] = [arr[j], arr[i]]); // Heap class, default maximum heap class Heap { constructor(cmp = defaultCmp) { this.container = []; this.cmp = cmp; } // insert insert(data) { const { container, cmp } = this; container.push(data); let index = this.size() - 1; while (index) { let parent = (index - 1) >> 1; if (!cmp(container[index], container[parent])) { return; } swap(container, index, parent); index = parent; } } // Eject the top of the heap and return pop() { const { container, cmp } = this; if (!this.size()) { return null; } swap(container, 0, this.size() - 1); const res = container.pop(); const length = this.size(); let index = 0, exchange = index * 2 + 1; while (exchange < length) { // //In the case of maximum heap: if there is a right node, and the value of the right node is greater than that of the left node let right = index * 2 + 2; if (right < length && cmp(container[right], container[exchange])) { exchange = right; } if (!cmp(container[exchange], container[index])) { break; } swap(container, exchange, index); index = exchange; exchange = index * 2 + 1; } return res; } // Get heap size size() { return this.container.length; } } const findMaximizedCapital = (k, w, profits, capital) => { const n = profits.length; const arr = new Array(n); // Combine capital and profit by item number [capital, net profit] for (let i = 0; i < n; i++) { arr[i] = [capital[i], profits[i]]; } // Rank projects by required capital from small to large arr.sort((a, b) => a[0] - b[0]); // Create maximum heap const maxHeap = new Heap(); let cur = 0; for (let i = 0; i < k; i++) { while (cur < n && arr[cur][0] <= w) { // Insert the profits from all qualified items into the heap maxHeap.insert(arr[cur++][1]); } if (maxHeap.size()) { // Heap is not empty // Take out the top of the pile, which is the maximum net profit and update your own capital w w += maxHeap.pop(); } else { // If the heap is empty, exit the loop directly // Because no items that meet the conditions have entered the heap break; } } return w; };

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