An investor is looking to maximize their capital by undertaking a set of profitable projects. Due to limited time and resources, they can complete at most k distinct projects.
There are n available projects. Each project i has:
The investor starts with an initial capital of c. After completing a project, its profit is immediately added to the investor's current capital.
The goal is to choose up to k different projects in a way that maximizes the investor’s final capital. Return the maximum capital achievable after completing these projects.
It is guaranteed that the answer fits within a 32-bit signed integer.
Constraints:
1≤ k ≤103
0≤ c ≤ 109
n== profits.length
n== capitals.length
1≤n≤103
0≤profits[i] ≤104
0≤ capitals[i] ≤109
You may have already brainstormed some approaches and have an idea of how to solve this problem. Let’s explore some of these approaches and figure out which one to follow based on considerations such as time complexity and implementation constraints.
The naive approach is to traverse every value of the capitals array based on the available capital. If the current capital is less than or equal to the capital value in the array, then store the profit value in a new array that corresponds to the capital index. Whenever the current capital becomes less than the capital value in the array, we’ll select the project with the largest profit value. The selected profit value will be added to the previous capital. Repeat this process until we get the required number of projects containing the maximum profit.
The time complexity of this solution is O(n2), where n represents the number of projects. This is because we need O(n) time to traverse the capitals array in order to find affordable projects. Additionally, for each subset of affordable projects, we would need another O(n) search to narrow down the project list to the one that yields the highest profit. The space complexity for storing the profit values in a new array is O(n).