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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.
1using System;
2using System.Collections.Generic;
3
4class Solution
5{
6 public int FindMaximizedCapital(int k, int w, int[] profits, int[] capital)
7 {
8 int n = profits.Length;
9 (int capital, int profit)[] projects = new (int, int)[n];
10 for (int i = 0; i < n; ++i)
11 {
12 projects[i] = (capital[i], profits[i]);
13 }
14 Array.Sort(projects, (a, b) => a.capital - b.capital);
var maxProfitHeap = new SortedSet<(int profit, int index)>();
int curr = 0;
while (k-- > 0)
{
while (curr < n && projects[curr].capital <= w)
{
maxProfitHeap.Add((projects[curr].profit, curr));
++curr;
}
if (maxProfitHeap.Count == 0) break;
var maxProfit = new List<(int profit, int index)>(maxProfitHeap);
var maxVal = maxProfit[^1];
w += maxVal.profit;
maxProfitHeap.Remove(maxVal);
}
return w;
}
}
class Program
{
static void Main(string[] args)
{
Solution sol = new Solution();
int k = 2, w = 0;
int[] profits = {1, 2, 3};
int[] capital = {0, 1, 1};
int result = sol.FindMaximizedCapital(k, w, profits, capital);
Console.WriteLine($"The final maximized capital is: {result}");
}
}This C# implementation leverages .NET's SortedSet to simulate a max-heap through sorting of potential profits. Following input project list creation and sorting, the algorithm iteratively checks for valid project execution based on current capital, and chooses the best profit-generating project repeatedly.
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.
1#include <stdio.h>
2#include <stdlib.h>
3#include <stdbool.h>
4
5int max(int a, int b) {
6 return a > b ? a : b;
7}
8
9int dp(int k, int w, int* profits, int* capital, int n, int current, int cache[][1001]) {
10 if (k == 0 || current == n) {
11 return w;
12 }
13
14 if (cache[current][w] != -1) {
15 return cache[current][w];
16 }
17
18 int notTake = dp(k, w, profits, capital, n, current + 1, cache);
19 int take = w >= capital[current] ? dp(k - 1, w + profits[current], profits, capital, n, current + 1, cache) : 0;
20 return cache[current][w] = max(take, notTake);
21}
22
23int findMaximizedCapital(int k, int w, int* profits, int* capital, int n) {
24 int cache[1000][1001];
25 for (int i = 0; i < 1000; ++i) {
26 for (int j = 0; j < 1001; ++j) {
27 cache[i][j] = -1;
28 }
29 }
30
31 return dp(k, w, profits, capital, n, 0, cache);
32}
33
34int main() {
35 int k = 2;
36 int w = 0;
37 int profits[] = {1, 2, 3};
38 int capital[] = {0, 1, 1};
39 int n = sizeof(profits) / sizeof(profits[0]);
40 int result = findMaximizedCapital(k, w, profits, capital, n);
41 printf("The final maximized capital is: %d\n", result);
42 return 0;
43}The C solution employs a recursive solution with dynamic programming utilizing memoization. Recursive strategy carefully reviews every set of projects within feasible conditions, referencing precomputed results when similar conditions occur.
The solution includes memoization cache for incremental computational cost alleviation over the choices possible across feasible capital, maximum projects to complete, consequently yielding comprehensive checks on potential project execution paths.