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This approach uses a greedy algorithm in conjunction with priority queues to select the projects maximizing profits. The core idea is to iteratively choose the most profitable project that can be initiated with the current available capital.
Initially, projects are stored in a list along with their capital and profit requirements. We prioritize by try fetching the most profitable projects. We maintain available projects in a max-heap (or priority queue) to achieve this efficiently.
We attempt to start at most k projects, updating the available capital every time we include a new project. For every iteration, once we identify the most feasible projects that can be executed given the current capital, we pick and execute the one with the highest profit and update our available capital. This process repeats until no more projects can be started or we've reached k projects.
Time Complexity: O(N log N + K log N), where N is the number of projects. This involves an initial sort and effectively K max operations on a priority-like structure.
Space Complexity: O(N) due to storage used for auxiliary lists and heaps.
1#include <iostream>
2#include <vector>
3#include <queue>
4#include <algorithm>
5
6using namespace std;
7
8class Solution {
9public:
10 int findMaximizedCapital(int k, int w, vector<int>& Profits, vector<int>& Capital) {
11 vector<pair<int, int>> projects;
12 for (int i = 0; i < Profits.size(); ++i) {
13 projects.push_back({Capital[i], Profits[i]});
14 }
sort(projects.begin(), projects.end());
priority_queue<int> maxProfitHeap;
int curr = 0;
while (k--) {
while (curr < projects.size() && projects[curr].first <= w) {
maxProfitHeap.push(projects[curr++].second);
}
if (maxProfitHeap.empty()) break;
w += maxProfitHeap.top();
maxProfitHeap.pop();
}
return w;
}
};
int main() {
int k = 2, w = 0;
vector<int> Profits{1, 2, 3};
vector<int> Capital{0, 1, 1};
Solution sol;
int result = sol.findMaximizedCapital(k, w, Profits, Capital);
cout << "The final maximized capital is: " << result << endl;
return 0;
}The C++ solution uses libraries like algorithm and priority_queue to efficiently manage collections and operations. Projects are sorted based on capital requirements, which allows efficient feasibility checks. A max-heap (using priority_queue) stores potential profits from feasible projects, which are iteratively selected for maximizing profit.
At each step up to <= k, the most profitable feasible project is selected, and its profit added to available capital.
This dynamic programming approach delineates with the constraints arising from the number of projects executable. We maintain an accruing structure documenting the maximum possible capital achievable for the number of projects done. The dynamic programming strategy opts for inherent flexibility where feasible, owing to recursive checks across all projects capable of initiating.
Memoization ensures repeated subproblem results are accessed without re-computation, improving overall efficiency particularly in scenarios involving expedient overlap of capital ranges per project.
Despite the computationally exhaustive exploration of all project combinations, dynamic memorized caching dramatically curtails avoidant efforts during feasibility analysis—gaining crucial strategic leverage with convenience in considering bounded project completion restrictions.
Time Complexity: O(N * W * K) — represents total options reviewed in DP tableau per states, characterized by projects count, capital capability, projects aimed for completion.
Space Complexity: O(N * W) — retains memory for tailored subproblems caching overseeing potential states across projects under capital explored known projects.
1def findMaximizedCapital(k, w, profits, capital):
2 n = len(profits)
3 projects = list(zip(capital, profits))
4 projects.sort()
5 dp = [[-1] * (k + 1) for _ in range(n)]
6
7 def maximum(k, w, idx):
8 if k == 0 or idx == n:
9 return w
10 if dp[idx][k] != -1:
11 return dp[idx][k]
12
13 notTake = maximum(k, w, idx + 1)
14 take = 0
15 if w >= projects[idx][0]:
16 take = maximum(k - 1, w + projects[idx][1], idx + 1)
17
18 dp[idx][k] = max(take, notTake)
19 return dp[idx][k]
20
21 return maximum(k, w, 0)
22
23if __name__ == '__main__':
24 k = 2
25 w = 0
26 profits = [1, 2, 3]
27 capital = [0, 1, 1]
28 result = findMaximizedCapital(k, w, profits, capital)
29 print(f"The final maximized capital is: {result}")Python implementation continues memorized logic in a recursive scheme, capturing potential resultant savings per states considering active capital and project tendency.
Recursion navigation ensures attempt pathways build upon pre-completed operations enumerated regarding projects long-lost and those advancing given adjacent states.