To maximize LeetCode's capital before its IPO, we need to select projects that provide the highest profit while ensuring we have enough initial capital to start them. By focusing on projects we can afford and that offer the best returns, we can increase our capital efficiently.
-
Initialization:
- Utilize two sorted sets:
availableProjects
: Contains projects sorted by their required capital.profitableProjects
: Contains projects sorted by their potential profit in descending order.
- Utilize two sorted sets:
-
Project Selection:
- Iterate up to
numberOfProjects
times:- Move all projects from
availableProjects
toprofitableProjects
that can be started with the current capital. - If no profitable projects are available, terminate the selection process.
- Select the most profitable project from
profitableProjects
, add its profit to the current capital, and remove it from the set.
- Move all projects from
- Iterate up to
-
Return the Result:
- Output the final capital after selecting up to the specified number of projects.
- Time complexity: O(n log n + k log n), where
n
is the number of projects andk
is the number of projects selected. This includes sorting and selecting projects based on profitability and capital. - Space complexity: O(n), primarily due to storing projects in the sorted sets.
public class FindMaximizedCapital
{
public int Solve(int numberOfProjects, int initialCapital, int[] profits, int[] capitals)
{
// Create a sorted set to store available projects by capital required
var availableProjects = new SortedSet<(int capital, int profit, int index)>();
// Create a sorted set to store profitable projects by profit
var profitableProjects = new SortedSet<(int profit, int index)>(
Comparer<(int profit, int index)>
.Create((a, b) => b.profit == a.profit ? a.index.CompareTo(b.index) : b.profit.CompareTo(a.profit)));
// Add all projects to the available projects set
for (int i = 0; i < profits.Length; i++)
{
availableProjects.Add((capitals[i], profits[i], i));
}
int currentCapital = initialCapital;
// Iterate up to the maximum number of projects we can perform
for (int i = 0; i < numberOfProjects; i++)
{
// Move projects that can be started with the current capital to the profitable projects set
while (availableProjects.Count > 0 && availableProjects.Min.capital <= currentCapital)
{
var project = availableProjects.Min;
availableProjects.Remove(project);
profitableProjects.Add((project.profit, project.index));
}
// If no profitable projects are available, break the loop
if (profitableProjects.Count == 0)
{
break;
}
// Select the most profitable project and increase current capital
var bestProject = profitableProjects.Min;
profitableProjects.Remove(bestProject);
currentCapital += bestProject.profit;
}
return currentCapital;
}
}