Solution For Ipo

Note: This solution assumes understanding of IPO problem statement and is for educational purposes only.

The IPO problem on LeetCode is a greedy algorithm problem that can be solved using a priority queue. The problem statement is as follows:

You are given a set of investment projects with their respective profits, and a way to split them into smaller projects. You are also given an initial capital starting from 0.

At each step, you can choose to invest in one of the available projects, with the goal of maximizing your total capital. However, you cannot invest in a project if your current capital is less than the project’s minimum capital required.

You are allowed to undertake multiple projects simultaneously, but the total initial cost of all the projects cannot exceed your available capital.

Write a function to find the maximum capital you can obtain.

The function definition is:
def findMaximizedCapital(k: int, w: int, profits: List[int], capital: List[int]) -> int:

The inputs are:
– k: the maximum number of projects you can undertake simultaneously.
– w: your initial capital.
– profits: a list of integers representing the profits of the investment projects.
– capital: a list of integers representing the minimum capital required to invest in each project.

The output should be the maximum capital you can obtain after investing in k projects.

The first step in solving this problem is to create a list of tuples for each project, where each tuple contains the project’s minimum capital required and its corresponding profit.

projects = [(c, p) for c, p in zip(capital, profits)]

Next, we sort the projects in order of increasing minimum capital required.

projects.sort()

To keep track of the available projects, we use two priority queues:
– a max heap (profits_heap) to keep track of the profits of available projects, and
– a min heap (capital_heap) to keep track of the minimum capital required for each project.

We start by adding all the projects with a minimum capital less than or equal to w to the profits_heap.

profits_heap = [] capital_heap = [] for i in range(len(projects)):
if projects[i][0] <= w:
heappush(profits_heap, -projects[i][1])
else:
heappush(capital_heap, projects[i])

Next, we iterate k times and at each iteration we choose the project with the highest profit from the profits_heap and add its profit to w. We then add all the projects that require capital less than or equal to w to the profits_heap.

for i in range(k):
if len(profits_heap) == 0:
break
p = -heappop(profits_heap)
w += p
while len(capital_heap) > 0 and capital_heap[0][0] <= w:
c, p = heappop(capital_heap)
heappush(profits_heap, -p)

Finally, we return w, which is the maximum capital we can obtain after investing in k projects.

return w

The time complexity of this solution is O(nlogn + klogn), where n is the number of projects and k is the number of projects we can undertake simultaneously. The space complexity is O(n).

Step by Step Implementation For Ipo

public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) {
    
    // create a min heap of projects with their profits
    PriorityQueue projects = new PriorityQueue<>(new Comparator() {
        public int compare(int[] a, int[] b) {
            return a[1] - b[1];
        }
    });
    
    // create a max heap of capitals
    PriorityQueue capitals = new PriorityQueue<>(new Comparator() {
        public int compare(Integer a, Integer b) {
            return b - a;
        }
    });
    
    // add all projects to the min heap
    for (int i = 0; i < Profits.length; i++) {
        projects.add(new int[]{Profits[i], Capital[i]});
    }
    
    // iterate through all projects
    for (int i = 0; i < k; i++) {
        
        // add all available capitals to the max heap
        while (!projects.isEmpty() && projects.peek()[1] <= W) {
            capitals.add(projects.poll()[0]);
        }
        
        // if there are no more available projects, break
        if (capitals.isEmpty()) {
            break;
        }
        
        // invest in the most profitable project
        W += capitals.poll();
    }
    
    return W;
}

def ipo(k, W, Profits, Capital): 
    
    # Initialize the result 
    res = 0
    
    # Create a priority queue 
    # to store live projects 
    pq = [] 
    
    # Insert the initial projects 
    # with their initial capital 
    for i in range(len(Profits)): 
        heapq.heappush(pq, (Capital[i], Profits[i], Capital[i])) 
    
    # While we have live projects 
    while pq: 
        
        # Extract the minimum capital 
        # project from the queue 
        _, curr_profit, curr_cap = heapq.heappop(pq) 
        
        # Check if the current project 
        # can be completed with the 
        # available initial capital. 
        # If not, then continue 
        if W >= curr_cap: 
            
            # Increase the result 
            res += curr_profit 
            
            # Decrease the available 
            # initial capital 
            W -= curr_cap 
            
            # For all live projects, 
            # if the available initial 
            # capital is more than 
            # the project capital, 
            # then insert the project 
            # in the priority queue 
            for i in range(len(Profits)): 
                if W >= Capital[i]: 
                    heapq.heappush(pq, (Capital[i], Profits[i], Capital[i])) 
                    
    return res

var ipo = function(p, c, n) {
    // p is the array of profits
    // c is the array of capital
    // n is the number of projects
    
    // we need to return the maximum profit that can be made
    
    // first, we sort the arrays by profit
    p.sort(function(a, b){return a-b});
    c.sort(function(a, b){return a-b});
    
    // next, we create a min heap with the capital array
    var heap = new minHeap(c);
    
    // finally, we iterate through the profit array
    // for each profit, we check if there is enough capital
    // to invest in that project
    // if there is, we update the heap and add the profit to our total
    var total = 0;
    for (var i = 0; i < p.length; i++) {
        if (heap.size() > 0 && heap.peek() <= p[i]) {
            total += heap.poll();
        }
        if (i < n) {
            heap.add(c[i]);
        }
    }
    
    return total;
};

1) Sort the projects in decreasing order of their profit-per-share ratio.

2) Select the first project and add it to the final list of projects.

3) For each remaining project, compute its profit-per-share ratio. If this ratio is greater than the profit-per-share ratio of the last project in the final list, then add the project to the final list.

4) Return the final list of projects.

using System; 
using System.Collections.Generic; 
using System.Linq; 

public class Solution { 
    public int FindMaximizedCapital(int k, int W, int[] Profits, int[] Capital) { 

// k is the number of times we can invest 
// W is our initial capital 
// Profits is an array of the potential profits we could make from investing in a certain company 
// Capital is an array of the amount of capital that a certain company requires in order to be invested in 

        // we need to track both the potential profits and the required capital in order to make the investment 
        var companies = new List>(); 

        for (int i = 0; i < Profits.Length; i++) { 
            companies.Add(Tuple.Create(Profits[i], Capital[i])); 
        } 

        // we need to sort the companies in order of increasing required capital 
        companies.Sort((x, y) => x.Item2.CompareTo(y.Item2)); 

        // we also need a min heap to keep track of the potential profits we could make from the companies we can invest in 
        var minHeap = new MinHeap(); 

        int currentCapital = W; 
        int currentProfit = 0; 

        // we go through the list of companies until we have either invested k times or we have no more companies to invest in 
        while (k > 0 && companies.Count > 0) { 

            // we only want to consider companies that we have enough capital to invest in 
            while (companies.Count > 0 && currentCapital >= companies[0].Item2) { 

                // we add the company to the min heap 
                minHeap.Add(companies[0].Item1); 

                // and remove it from the list of companies 
                companies.RemoveAt(0); 
            } 

            // if we have companies to invest in, we make our investment 
            if (minHeap.Count > 0) { 
                currentProfit += minHeap.Remove(); 
                k--; 
            } 

            // we update our current capital after making our investment 
            currentCapital += currentProfit; 
            currentProfit = 0; 
        } 

        return currentCapital; 
    } 
} 

// a min heap implementation 
public class MinHeap { 
    private List heap; 

    public MinHeap() { 
        heap = new List(); 
    } 

    public void Add(int element) { 

// we add the element to the end of the list 
        heap.Add(element); 

// and then heapify up 
        HeapifyUp(heap.Count - 1); 
    } 

    public int Remove() { 

// we remove the first element in the heap (the minimum element) 
        int removedElement = heap[0]; 

// and replace it with the last element in the list 
        heap[0] = heap[heap.Count - 1]; 
        heap.RemoveAt(heap.Count - 1); 

// then heapify down 
        HeapifyDown(0); 

        return removedElement; 
    } 

    private void HeapifyUp(int index) { 

// we compare the element at the given index with its parent 
        int parentIndex = (index - 1) / 2; 

// if the element is smaller than its parent, we swap them 
        if (heap[index] < heap[parentIndex]) { 
            Swap(index, parentIndex); 

// and heapify up from the parent's index 
            HeapifyUp(parentIndex); 
        } 
    } 

    private void HeapifyDown(int index) { 

// we compare the element at the given index with its children 
        int leftChildIndex = 2 * index + 1; 
        int rightChildIndex = 2 * index + 2; 

// if the left child exists and is smaller than the element, we swap them 
        if (leftChildIndex < heap.Count && heap[leftChildIndex] < heap[index]) { 
            Swap(leftChildIndex, index); 

// and heapify down from the left child's index 
            HeapifyDown(leftChildIndex); 
        } 

// if the right child exists and is smaller than the element, we swap them 
        if (rightChildIndex < heap.Count && heap[rightChildIndex] < heap[index]) { 
            Swap(rightChildIndex, index); 

// and heapify down from the right child's index 
            HeapifyDown(rightChildIndex); 
        } 
    } 

// a helper function to swap two elements in the heap 
    private void Swap(int index1, int index2) { 
        int temp = heap[index1]; 
        heap[index1] = heap[index2]; 
        heap[index2] = temp; 
    } 
}