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Code Implementation

Capital Investment Optimizer Implementation

Below is the implementation of the capital investment optimizer:

solution.js

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/**
 * Brute Force Approach
 * Time Complexity: O(k * n) - For each of the k projects, we scan all n projects
 * Space Complexity: O(n) - We need space to track completed projects
 *
 * @param {number} k - Maximum number of projects
 * @param {number} w - Initial capital
 * @param {number[]} profits - Array of project profits
 * @param {number[]} capital - Array of project capital requirements
 * @return {number} - Maximum final capital
 */
function findMaximizedCapitalBruteForce(k, w, profits, capital) {
  let currentCapital = w;
  const completed = new Set();
 
  for (let i = 0; i < k; i++) {
    let maxProfit = 0;
    let maxProfitIndex = -1;
 
    // Find the project with the highest profit that we can start
    for (let j = 0; j < profits.length; j++) {
      if (!completed.has(j) && capital[j] <= currentCapital && profits[j] > maxProfit) {
        maxProfit = profits[j];
        maxProfitIndex = j;
      }
    }
 
    // If no project can be started, break
    if (maxProfitIndex === -1) {
      break;
    }
 
    // Complete the project and update capital
    currentCapital += maxProfit;
    completed.add(maxProfitIndex);
  }
 
  return currentCapital;
}

Step-by-Step Explanation

Let's break down the implementation:

1. Understand the Problem: First, understand that we need to find the maximum capital we can accumulate by completing at most k projects, where each project requires a minimum capital to start and yields a profit upon completion.
2. Implement the Brute Force Approach: For each step, scan all projects to find the one with the highest profit that can be started with the current capital.
3. Optimize with Two Heaps: Use a min heap to sort projects by capital requirements and a max heap to sort available projects by profit.
4. Handle Edge Cases: Consider edge cases such as when k is large, when the initial capital is very small or very large, or when there are no projects that can be started.
5. Analyze Time and Space Complexity: Understand the time and space complexity of each approach to choose the most efficient one for the given constraints.
6. Test the Solution: Verify the solution with the provided examples and additional test cases.

Previous Next

String Problems

Palindrome Checker

Determine if a string reads the same forward and backward.

EasyString Manipulation

Message Encryption

Reverse a string to create a simple encryption system.

EasyString Manipulation

Word Puzzle Challenge

Check if two strings are anagrams of each other.

EasyString Manipulation

Learning Objective

Implement the capital investment optimizer solution in different programming languages.

Capital Investment Optimizer Implementation

Below is the implementation of the capital investment optimizer in different programming languages. Select a language tab to view the corresponding code.

solution.js

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/**
 * Brute Force Approach
 * Time Complexity: O(k * n) - For each of the k projects, we scan all n projects
 * Space Complexity: O(n) - We need space to track completed projects
 *
 * @param {number} k - Maximum number of projects
 * @param {number} w - Initial capital
 * @param {number[]} profits - Array of project profits
 * @param {number[]} capital - Array of project capital requirements
 * @return {number} - Maximum final capital
 */
function findMaximizedCapitalBruteForce(k, w, profits, capital) {
  let currentCapital = w;
  const completed = new Set();
 
  for (let i = 0; i < k; i++) {
    let maxProfit = 0;
    let maxProfitIndex = -1;
 
    // Find the project with the highest profit that we can start
    for (let j = 0; j < profits.length; j++) {
      if (!completed.has(j) && capital[j] <= currentCapital && profits[j] > maxProfit) {
        maxProfit = profits[j];
        maxProfitIndex = j;
      }
    }
 
    // If no project can be started, break
    if (maxProfitIndex === -1) {
      break;
    }
 
    // Complete the project and update capital
    currentCapital += maxProfit;
    completed.add(maxProfitIndex);
  }
 
  return currentCapital;
}

Step-by-Step Explanation

11. Understand the Problem

First, understand that we need to find the maximum capital we can accumulate by completing at most k projects, where each project requires a minimum capital to start and yields a profit upon completion.

22. Implement the Brute Force Approach

For each step, scan all projects to find the one with the highest profit that can be started with the current capital.

33. Optimize with Two Heaps

Use a min heap to sort projects by capital requirements and a max heap to sort available projects by profit.

44. Handle Edge Cases

Consider edge cases such as when k is large, when the initial capital is very small or very large, or when there are no projects that can be started.

55. Analyze Time and Space Complexity

Understand the time and space complexity of each approach to choose the most efficient one for the given constraints.

66. Test the Solution

Verify the solution with the provided examples and additional test cases.

Language-Specific Notes

javascript

JavaScript doesn't have built-in heap data structures, so we need to implement them or use libraries.
The Set data structure is used to efficiently track completed projects in the brute force approach.
JavaScript's sort() method can be used to sort arrays, but it's less efficient than using heaps for this problem.

python

Python's heapq module provides heap queue algorithm functions, which implement a min heap.
For a max heap, we can negate the values when pushing and popping from the heap.
The zip function is used to pair capital and profit values for easier processing.

java

Java's PriorityQueue class implements a min heap by default.
For a max heap, we can use Collections.reverseOrder() as the comparator.
Arrays are used to store pairs of capital and profit values in the min heap.

cpp

C++'s std::priority_queue implements a max heap by default.
For a min heap, we need to use std::greater as the comparator.
std::pair is used to store pairs of capital and profit values in the min heap.

Key Takeaways

The greedy approach of always choosing the project with the highest profit among available projects is optimal for this problem.
Using two heaps (priority queues) significantly improves the efficiency of finding available projects and the one with the highest profit.
The problem demonstrates the importance of efficiently tracking and updating available options as conditions change.
This is a classic example of a resource allocation problem where we need to make optimal decisions with limited resources.
The solution pattern can be applied to other problems involving sequential decision-making with changing constraints.
Understanding when to use different data structures (like heaps) can lead to significant performance improvements.

Try It Yourself

Modify the code to implement an alternative approach and test it with the same examples.

Challenge:

Implement a function that solves the capital investment optimizer problem using a different approach than shown above.

Common Edge Cases

No Projects Can Be Started

If the initial capital is too low to start any project, the final capital will be the same as the initial capital.

k = 3, w = 0, profits = [1,2,3], capital = [1,2,3] → 0

All Projects Can Be Started

If the initial capital is high enough to start all projects, we should choose the k projects with the highest profits.

k = 2, w = 10, profits = [1,2,3], capital = [1,2,3] → 15

Fewer Than k Projects Available

If there are fewer than k projects available, we should complete all available projects.

k = 5, w = 0, profits = [1,2,3], capital = [0,1,2] → 6

Large k Value

If k is very large, we might end up completing all projects before reaching k.

k = 1000, w = 0, profits = [1,2,3], capital = [0,1,2] → 6

Solution Approach All Problems

Problem Solution Code

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Capital Investment Optimizer - Implementation

Code Implementation

Capital Investment Optimizer Implementation

Below is the implementation of the capital investment optimizer:

solution.js

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/**
 * Brute Force Approach
 * Time Complexity: O(k * n) - For each of the k projects, we scan all n projects
 * Space Complexity: O(n) - We need space to track completed projects
 *
 * @param {number} k - Maximum number of projects
 * @param {number} w - Initial capital
 * @param {number[]} profits - Array of project profits
 * @param {number[]} capital - Array of project capital requirements
 * @return {number} - Maximum final capital
 */
function findMaximizedCapitalBruteForce(k, w, profits, capital) {
  let currentCapital = w;
  const completed = new Set();
 
  for (let i = 0; i < k; i++) {
    let maxProfit = 0;
    let maxProfitIndex = -1;
 
    // Find the project with the highest profit that we can start
    for (let j = 0; j < profits.length; j++) {
      if (!completed.has(j) && capital[j] <= currentCapital && profits[j] > maxProfit) {
        maxProfit = profits[j];
        maxProfitIndex = j;
      }
    }
 
    // If no project can be started, break
    if (maxProfitIndex === -1) {
      break;
    }
 
    // Complete the project and update capital
    currentCapital += maxProfit;
    completed.add(maxProfitIndex);
  }
 
  return currentCapital;
}

Step-by-Step Explanation

Let's break down the implementation:

1. Understand the Problem: First, understand that we need to find the maximum capital we can accumulate by completing at most k projects, where each project requires a minimum capital to start and yields a profit upon completion.
2. Implement the Brute Force Approach: For each step, scan all projects to find the one with the highest profit that can be started with the current capital.
3. Optimize with Two Heaps: Use a min heap to sort projects by capital requirements and a max heap to sort available projects by profit.
4. Handle Edge Cases: Consider edge cases such as when k is large, when the initial capital is very small or very large, or when there are no projects that can be started.
5. Analyze Time and Space Complexity: Understand the time and space complexity of each approach to choose the most efficient one for the given constraints.
6. Test the Solution: Verify the solution with the provided examples and additional test cases.

Previous Next

String Problems

Palindrome Checker

Determine if a string reads the same forward and backward.

EasyString Manipulation

Message Encryption

Reverse a string to create a simple encryption system.

EasyString Manipulation

Word Puzzle Challenge

Check if two strings are anagrams of each other.

EasyString Manipulation

Learning Objective

Implement the capital investment optimizer solution in different programming languages.

Capital Investment Optimizer Implementation

Below is the implementation of the capital investment optimizer in different programming languages. Select a language tab to view the corresponding code.

solution.js

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23
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/**
 * Brute Force Approach
 * Time Complexity: O(k * n) - For each of the k projects, we scan all n projects
 * Space Complexity: O(n) - We need space to track completed projects
 *
 * @param {number} k - Maximum number of projects
 * @param {number} w - Initial capital
 * @param {number[]} profits - Array of project profits
 * @param {number[]} capital - Array of project capital requirements
 * @return {number} - Maximum final capital
 */
function findMaximizedCapitalBruteForce(k, w, profits, capital) {
  let currentCapital = w;
  const completed = new Set();
 
  for (let i = 0; i < k; i++) {
    let maxProfit = 0;
    let maxProfitIndex = -1;
 
    // Find the project with the highest profit that we can start
    for (let j = 0; j < profits.length; j++) {
      if (!completed.has(j) && capital[j] <= currentCapital && profits[j] > maxProfit) {
        maxProfit = profits[j];
        maxProfitIndex = j;
      }
    }
 
    // If no project can be started, break
    if (maxProfitIndex === -1) {
      break;
    }
 
    // Complete the project and update capital
    currentCapital += maxProfit;
    completed.add(maxProfitIndex);
  }
 
  return currentCapital;
}

Step-by-Step Explanation

11. Understand the Problem

22. Implement the Brute Force Approach

For each step, scan all projects to find the one with the highest profit that can be started with the current capital.

33. Optimize with Two Heaps

Use a min heap to sort projects by capital requirements and a max heap to sort available projects by profit.

44. Handle Edge Cases

Consider edge cases such as when k is large, when the initial capital is very small or very large, or when there are no projects that can be started.

55. Analyze Time and Space Complexity

Understand the time and space complexity of each approach to choose the most efficient one for the given constraints.

66. Test the Solution

Verify the solution with the provided examples and additional test cases.

Language-Specific Notes

javascript

JavaScript doesn't have built-in heap data structures, so we need to implement them or use libraries.
The Set data structure is used to efficiently track completed projects in the brute force approach.
JavaScript's sort() method can be used to sort arrays, but it's less efficient than using heaps for this problem.

python

Python's heapq module provides heap queue algorithm functions, which implement a min heap.
For a max heap, we can negate the values when pushing and popping from the heap.
The zip function is used to pair capital and profit values for easier processing.

java

Java's PriorityQueue class implements a min heap by default.
For a max heap, we can use Collections.reverseOrder() as the comparator.
Arrays are used to store pairs of capital and profit values in the min heap.

cpp

C++'s std::priority_queue implements a max heap by default.
For a min heap, we need to use std::greater as the comparator.
std::pair is used to store pairs of capital and profit values in the min heap.

Key Takeaways

The greedy approach of always choosing the project with the highest profit among available projects is optimal for this problem.
Using two heaps (priority queues) significantly improves the efficiency of finding available projects and the one with the highest profit.
The problem demonstrates the importance of efficiently tracking and updating available options as conditions change.
This is a classic example of a resource allocation problem where we need to make optimal decisions with limited resources.
The solution pattern can be applied to other problems involving sequential decision-making with changing constraints.
Understanding when to use different data structures (like heaps) can lead to significant performance improvements.

Try It Yourself

Modify the code to implement an alternative approach and test it with the same examples.

Challenge:

Implement a function that solves the capital investment optimizer problem using a different approach than shown above.

Common Edge Cases

No Projects Can Be Started

If the initial capital is too low to start any project, the final capital will be the same as the initial capital.

k = 3, w = 0, profits = [1,2,3], capital = [1,2,3] → 0

All Projects Can Be Started

If the initial capital is high enough to start all projects, we should choose the k projects with the highest profits.

k = 2, w = 10, profits = [1,2,3], capital = [1,2,3] → 15

Fewer Than k Projects Available

If there are fewer than k projects available, we should complete all available projects.

k = 5, w = 0, profits = [1,2,3], capital = [0,1,2] → 6

Large k Value

If k is very large, we might end up completing all projects before reaching k.

k = 1000, w = 0, profits = [1,2,3], capital = [0,1,2] → 6

Solution Approach All Problems

Code Implementation

Capital Investment Optimizer Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Capital Investment Optimizer Implementation

Step-by-Step Explanation

11. Understand the Problem

22. Implement the Brute Force Approach

33. Optimize with Two Heaps

44. Handle Edge Cases

55. Analyze Time and Space Complexity

66. Test the Solution

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

No Projects Can Be Started

All Projects Can Be Started

Fewer Than k Projects Available

Large k Value

Capital Investment Optimizer - Implementation

Code Implementation

Capital Investment Optimizer Implementation

Step-by-Step Explanation

String ProblemsShow All

Palindrome Checker

Message Encryption

Word Puzzle Challenge

Learning Objective

Capital Investment Optimizer Implementation

Step-by-Step Explanation

11. Understand the Problem

22. Implement the Brute Force Approach

33. Optimize with Two Heaps

44. Handle Edge Cases

55. Analyze Time and Space Complexity

66. Test the Solution

Language-Specific Notes

javascript

python

java

cpp

Key Takeaways

Try It Yourself

Challenge:

Common Edge Cases

No Projects Can Be Started

All Projects Can Be Started

Fewer Than k Projects Available

Large k Value

String Problems

String Problems