101 Logo
onenoughtone

Solution Approach

Solution Overview

There are 2 main approaches to solve this problem:

  1. Brute Force Approach - Complex approach
  2. Two Heaps Approach - Complex approach

Approach 1: Brute Force Approach

Let's start by understanding the problem: we need to find the maximum capital we can accumulate by completing at most k projects, where each project requires a minimum capital to start and yields a profit upon completion.

Thinking Process: The most straightforward approach would be to try all possible combinations of at most k projects and find the one that maximizes the final capital. However, this would be extremely inefficient as there are potentially exponentially many combinations.

A slightly better brute force approach would be to simulate the process: at each step, we consider all available projects (those with required capital less than or equal to our current capital), choose the one with the highest profit, complete it, update our capital, and repeat until we've completed k projects or there are no more available projects.

Intuition: This approach directly follows the problem statement. We greedily choose the project with the highest profit among all available projects at each step. However, the inefficiency comes from having to scan all projects at each step to find the available ones and the one with the highest profit.

Algorithm:

  1. Initialize the current capital to w.
  2. Repeat the following steps at most k times or until no more projects can be started:
  3. a. Find all projects that can be started with the current capital.
  4. b. If no such projects exist, break the loop.
  5. c. Among these projects, find the one with the highest profit.
  6. d. Complete this project, add its profit to the current capital, and mark it as completed.
  7. Return the final capital.

Time Complexity:

O(k * n)

Where k is the maximum number of projects and n is the total number of projects. In the worst case, we need to scan all n projects k times to find the one with the highest profit.

Space Complexity:

O(n)

We need O(n) space to keep track of which projects have been completed.

Approach 2: Two Heaps Approach

We can optimize the brute force approach by using two priority queues (heaps) to efficiently track available projects and find the one with the highest profit.

Thinking Process: The key insight is that we need to efficiently track two things:

  1. All projects sorted by their required capital (to quickly find which projects become available as our capital increases).
  2. Available projects sorted by their profit (to quickly find the project with the highest profit).

We can use a min heap to sort projects by their required capital and a max heap to sort available projects by their profit. At each step, we move all newly available projects from the min heap to the max heap, then choose the project with the highest profit from the max heap.

Intuition: By using two heaps, we avoid scanning all projects at each step. The min heap allows us to efficiently find projects that become available as our capital increases, and the max heap allows us to efficiently find the project with the highest profit among all available projects.

Algorithm:

  1. Create a min heap (capitalHeap) to store all projects sorted by their required capital.
  2. Create a max heap (profitHeap) to store available projects sorted by their profit.
  3. Initialize the current capital to w.
  4. Add all projects to the capitalHeap.
  5. Repeat the following steps at most k times or until both heaps are empty:
  6. a. Move all projects from the capitalHeap to the profitHeap if their required capital is less than or equal to the current capital.
  7. b. If the profitHeap is empty, break the loop.
  8. c. Remove the project with the highest profit from the profitHeap and add its profit to the current capital.
  9. Return the final capital.

Time Complexity:

O(n log n)

Where n is the total number of projects. Adding all projects to the capitalHeap takes O(n log n) time. In the worst case, we might move each project from the capitalHeap to the profitHeap once, which takes O(n log n) time. Removing at most k projects from the profitHeap takes O(k log n) time. Since k ≤ n, the overall time complexity is O(n log n).

Space Complexity:

O(n)

We need O(n) space to store all projects in the heaps.

Pseudocode

solution.pseudo
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
function findMaximizedCapital(k, w, profits, capital):
    currentCapital = w
    completed = new Set()
    
    for i from 1 to k:
        maxProfit = 0
        maxProfitIndex = -1
        
        for j from 0 to profits.length - 1:
            if j not in completed and capital[j] <= currentCapital and profits[j] > maxProfit:
                maxProfit = profits[j]
                maxProfitIndex = j
        
        if maxProfitIndex == -1:
            break
        
        currentCapital += maxProfit
        completed.add(maxProfitIndex)
    
    return currentCapital
PreviousNext
ProblemSolutionCode
101 Logo
onenoughtone