LeetCode 520 LintCode 1202 IPO

Table of contents

Description

Suppose LeetCode will start its IPO 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 IPO. Since it has limited resources, it can only finish at most k distinct projects before the IPO. Help LeetCode design the best way to maximize its total capital after finishing at most k distinct projects.

You are given n projects where the ith project has a pure profit profits[i] and a minimum capital of capital[i] is needed to start it.

Initially, you have w capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.

Pick a list of at most k distinct projects from given projects to maximize your final capital, and return the final maximized capital.

The answer is guaranteed to fit in a 32-bit signed integer.

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 the project indexed 0.
After finishing it you will obtain profit 1 and your capital becomes 1.
With capital 1, you can either start the project indexed 1 or the project indexed 2.
Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.
Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.

Example 2:

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

Constraints:

• 1 <= k <= 10^5
• 0 <= w <= 10^9
• n == profits.length
• n == capital.length
• 1 <= n <= 10^5
• 0 <= profits[i] <= 10^4
• 0 <= capital[i] <= 10^9

Idea

Each time we try to pick a project, we can greedily perform the following.

1. Find all projects with capital less than or equal to the current capital w.
2. Pick the project with the maximum profit from the list of projects in step above. update the capital. we could use a max heap to keep the pending projects with cost that we could afford so it would be convenient to pick the maximum from that collection.

After k iterations, we return the capital at the end.

Complexity: Time $O(n \log n+k \log n)$, Space $O(n)$.

Python

class Solution:
    """247 ms, 43.00 mb"""

    def findMaximizedCapital(self, k: int, w: int, profits: list[int], capital: list[int]) -> int:
        pq, i = [], 0
        p = sorted(zip(capital, profits))
        for _ in range(k):
            while i < len(p) and p[i][0] <= w:
                heappush(pq, -p[i][1])
                i += 1
            if pq:
                w -= heappop(pq)
            else:  # stuck, no more projects with capital <= w
                break
        return w

Java

// 104 ms, 59.19 mb
class Solution {
    public int findMaximizedCapital(int k, int w, int[] profits, int[] capital) {
        int n = profits.length;
        ArrayList<int[]> p = new ArrayList<>(n);
        for (int i = 0; i < n; i++) p.add(new int[]{capital[i], profits[i]});
        p.sort(Comparator.comparingInt(a -> a[0]));
        PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.reverseOrder());
        int i = 0;
        while (k-- > 0) {
            while (i < n && p.get(i)[0] <= w) pq.add(p.get(i++)[1]);
            if (pq.isEmpty()) break;
            else w += pq.remove();
        }
        return w;
    }
}