# 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 PriorityQueueprojects = 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; } }