IPO
The problem can be solved using a greedy approach. We will use two data structures, a minheap to sort projects by capital and a maxheap to sort projects by profit.
 We will create a list of tuples where each tuple contains the capital and profit of a project.
 Sort this list by capital.
 Create a maxheap to store available projects based on profits.
 Iterate from 1 to k, at each step: a. Add all projects that can be started with current capital to the maxheap. b. Pick the project with the highest profit from the maxheap and add it to the capital.
 Return the final capital.
Here’s the code:


The solution has time complexity O(n log n) due to sorting and heap operations, and space complexity O(n). It returns the maximized capital after completing at most k projects.
Identifying Problem Isomorphism
“IPO” shares similarities with “Maximum Number of Eaten Apples”. Both use the concept of priority queues and greedy algorithms.
In “IPO”, you are given several projects, each with a certain profit and a minimum capital requirement to start. Starting with a certain amount of capital, you are to maximize your total capital by completing at most k distinct projects. The priority queue here helps you to keep track of the projects that you can currently undertake.
In “Maximum Number of Eaten Apples”, there is a tree yielding apples with a certain lifespan. Each day, you can eat one apple from any tree which has apples, but the apples decay after a certain number of days. Your task is to maximize the number of apples you can eat. The priority queue in this problem is used to keep track of apples about to decay, helping you to decide which apple to eat.
Both problems have a concept of managing resources over time to achieve the maximum benefit, thus, they are related in a way. However, these problems also have differences, such as the need to manage projects in “IPO”, and the concept of decaying resources in “Maximum Number of Eaten Apples”.
The problem “IPO” is more complex due to the multiple factors that need to be managed at the same time (profit, capital, and number of projects). In contrast, “Maximum Number of Eaten Apples” is a bit simpler as it only involves the lifespan and quantity of apples.
10 Prerequisite LeetCode Problems
The “IPO” problem involves using a priority queue (or heap data structure) to solve a greedy algorithm problem. It requires an understanding of concepts related to sorting, priority queues, and handling multiparameter objects.
Here are 10 problems to build up to solving the “IPO” problem:
“Top K Frequent Elements” (LeetCode Problem #347): This problem introduces the concept of a priority queue and how it can be used to solve problems.
“Kth Largest Element in an Array” (LeetCode Problem #215): This problem provides practice with heaps in finding the kth largest element, similar to selecting k profitable projects in the “IPO” problem.
“Sort List” (LeetCode Problem #148): This problem involves sorting a linked list, but the sorting concept can be applied to any type of data.
“Binary Heap Operations” (LeetCode Problem #1405): This problem introduces the basics of heap operations.
“Assign Cookies” (LeetCode Problem #455): This problem introduces a similar greedy approach used in the “IPO” problem.
“Best Time to Buy and Sell Stock II” (LeetCode Problem #122): This problem helps to understand the greedy approach in a simpler context.
“Meeting Rooms II” (LeetCode Problem #253): This problem introduces scheduling, which is similar to the “IPO” problem.
“Task Scheduler” (LeetCode Problem #621): This problem involves scheduling tasks based on priority, similar to scheduling projects in the “IPO” problem.
“Container With Most Water” (LeetCode Problem #11): This problem helps to understand the concept of maximizing profit or outcome based on two parameters, similar to selecting projects based on profit and capital in the “IPO” problem.
“Merge Intervals” (LeetCode Problem #56): This problem requires understanding of sorting and manipulating multiparameter objects.


Language Agnostic Coding Drills
This Python code is implementing a Greedy approach with the use of a Priority Queue to solve a problem of maximizing capital. Here are the concepts used in this code:
Defining a class and methods: The main structure of this code is a class
Solution
with a methodfindMaximizedCapital
. It’s a typical pattern of coding problems from platforms like LeetCode. The method is defined with several parameters and returns an integer.Working with lists: This code uses Python lists to store profits and capitals of different projects. Understanding how to create, access, and manipulate lists is essential.
List comprehensions: The code uses a list comprehension to create a list of tuples, each of which represents a project with its capital and profit.
Sorting: The
sort()
function is used to sort the list of projects based on their capitals. This step is critical in a greedy approach.Loops and conditions: The while loop is used to go through all the projects and add suitable projects into a priority queue. There’s also an if condition inside the loop to break it when there’s no suitable project.
Using a heap or priority queue: A heap is a special treebased data structure, and it’s used here as a priority queue to always fetch the project with the most profit. Python’s
heapq
module provides functions for creating minheap, but with a little trick (store the negative profit), it can act like a maxheap.Updating variables: The code updates variables
w
andk
inside the loop, which involves understanding variable assignment and arithmetic operations.
Here’s a stepbystep approach to solve this problem:
 Define the function with appropriate parameters.
 Create a list of tuples, where each tuple contains capital and profit of a project. Then sort the list based on capital.
 Initialize a heap to store the profits of projects that can be completed with the current capital.
 Start a loop for as long as there are attempts left to complete a project.
 Inside the loop, keep adding the projects that can be completed with the current capital to the heap.
 If there are no suitable projects in the heap, break the loop. Otherwise, pick the project with the maximum profit, complete it and add the profit to the total capital.
 Continue this process until you have used all attempts or there are no suitable projects left.
 Finally, return the maximum capital you can accumulate.
Once these concepts are understood and drills practiced, they can be combined to solve this problem.
Targeted Drills in Python
Defining a class and methods: Implement a class
Person
with methodsgreeting
which prints a hello message andintroduction
which introduces the person given thename
andage
.1 2 3 4 5 6
class Person: def greeting(self): print("Hello!") def introduction(self, name, age): print(f"My name is {name} and I am {age} years old.")
Working with lists: Given a list
[1, 2, 3, 4, 5]
, write a Python script to add 10 to each element in the list and print the updated list.1 2 3
numbers = [1, 2, 3, 4, 5] updated_numbers = [num+10 for num in numbers] print(updated_numbers)
List comprehensions: Given two lists
capital = [5, 10, 15]
andprofits = [20, 30, 40]
, create a list of tuples where each tuple contains a capital and corresponding profit.1 2 3 4
capital = [5, 10, 15] profits = [20, 30, 40] projects = [(c, p) for c, p in zip(capital, profits)] print(projects)
Sorting: Sort the
projects
list from the previous drill based on capital.1 2
projects.sort() print(projects)
Loops and conditions: Given a list of numbers from 1 to 10, use a loop to find all numbers that are even and print them out. Break the loop once you find 5 even numbers.
1 2 3 4 5 6 7 8
numbers = list(range(1, 11)) count = 0 for num in numbers: if num % 2 == 0: print(num) count += 1 if count == 5: break
Using a heap or priority queue: Create a minheap with numbers from 1 to 5. Then pop the smallest element and print it. Repeat the operation until the heap is empty.
1 2 3 4 5
import heapq heap = [5, 4, 3, 2, 1] heapq.heapify(heap) while heap: print(heapq.heappop(heap))
Updating variables: Initialize a variable
total = 0
. Then, in a loop from 1 to 5, add each number tototal
and print the updatedtotal
.1 2 3 4
total = 0 for i in range(1, 6): total += i print(total)
Once you are comfortable with these individual drills, you can try combining them to solve the final problem.