LeetCode 502: IPO Solution in Python – A Step-by-Step Guide

What Is LeetCode 502: IPO?

In LeetCode 502: IPO, you’re given an initial capital w, a number of projects you can choose (k), and two lists: profits (pure profit per project) and capital (cost to start each project). Your goal is to pick up to k projects in any order to maximize your final capital, but you can only start a project if your current capital meets or exceeds its cost. For example, with w=0, k=1, profits=[1,2,3], and capital=[0,1,1], you’d pick the project with profit 1 (cost 0), ending with capital 1. This problem tests your ability to prioritize under constraints, building on greedy concepts from LeetCode 455: Assign Cookies.

Problem Statement

• Input:
• k: max number of projects (integer).
• w: initial capital (integer).
• profits: list of profits per project.
• capital: list of costs per project.
• Output: Integer—maximum capital after picking up to k projects.
• Rules: Can only pick a project if current capital ≥ its cost; pick at most k projects.

Constraints

• 1 <= k <= 10⁵
• 0 <= w <= 10⁹
• profits.length == capital.length (1 to 10⁵)
• 0 <= profits[i], capital[i] <= 10⁹

Examples

• Input: k=2, w=0, profits=[1,2,3], capital=[0,1,1]
• Output: 4
• Pick profit 1 (cost 0), then profit 3 (cost 1).
• Input: k=1, w=0, profits=[1,2,3], capital=[1,1,1]
• Output: 0
• Can’t afford any project.
• Input: k=3, w=0, profits=[1,2,3], capital=[0,1,2]
• Output: 6
• Pick all three in order.

Understanding the Problem: Maximizing Capital Greedily

To solve LeetCode 502: IPO in Python, we need a strategy to pick up to k projects that boost our capital the most, given our starting w and the cost constraint. A naive approach might try all combinations, but with up to 10⁵ projects, that’s impractical. Instead, we’ll use:

• Best Solution (Heaps): O(n log n + k log n) time, O(n) space—greedy and efficient.
• Alternative Solution (Brute Force with Sorting): O(n log n + k * n) time, O(n) space—simpler but slower.

Let’s jump into the heap-based solution with a friendly breakdown!

Best Solution: Using Heaps for Greedy Selection

Why Heaps Are the Best Choice

The heap-based solution shines for LeetCode 502 because it’s both fast and smart. We use a min-heap for project costs (to find affordable ones) and a max-heap for profits (to pick the best). Time complexity is O(n log n + k log n)—sorting costs initially, then heap operations—and space is O(n) for the heaps. It’s like having a financial advisor who quickly finds affordable projects and picks the most lucrative ones!

How It Works

Think of this as a two-step investment plan:

• Step 1: Sort by Cost:
• Pair each project’s cost and profit, sort by cost.
• Use a min-heap to track affordable projects based on current capital.
• Step 2: Pick Greedily:
• While you can pick more projects (k > 0) and have affordable options:
• Pop all projects you can afford into a max-heap of profits.
• Pick the highest profit, add it to capital, repeat.
• Why It Works:
• Greedy choice: Always take the max profit from what you can afford.
• Heaps keep operations fast (O(log n) per push/pop).

It’s like cherry-picking the best deals as your wallet grows!

Step-by-Step Example

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

• Init:
• Projects: [(0,1), (1,2), (1,3)] (cost, profit pairs).
• Min-heap (costs): [(0,1), (1,2), (1,3)].
• Max-heap (profits): [].
• Capital w = 0, k = 2.
• Step 1:
• Pop (0,1) (cost ≤ 0), push 1 to max-heap.
• Max-heap: [-1].
• Step 2:
• Pop 1, w = 0 + 1 = 1, k = 1.
• Step 3:
• Pop (1,2) and (1,3) (cost ≤ 1), push 2 and 3 to max-heap.
• Max-heap: [-3, -2].
• Step 4:
• Pop 3, w = 1 + 3 = 4, k = 0.
• Result: 4.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down for clarity:

Copy codeimport heapq

class Solution:
    def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
        # Step 1: Pair projects and sort by capital
        projects = list(zip(capital, profits))  # [(cost, profit), ...]
        heapq.heapify(projects)  # Min-heap by cost

        # Step 2: Max-heap for profits
        available = []  # Max-heap (use negative for max)

        # Step 3: Pick up to k projects
        for _ in range(k):
            # Move affordable projects to max-heap
            while projects and projects[0][0] <= w:
                cost, profit = heapq.heappop(projects)
                heapq.heappush(available, -profit)  # Negative for max-heap

            # If no projects available, stop
            if not available:
                break

            # Pick the most profitable
            w += -heapq.heappop(available)  # Add profit (negate back)

        return w

• Line 5: Zip capital and profits into pairs, make a min-heap by cost.
• Line 8: Empty list for max-heap (profits).
• Line 11: Loop up to k times.
• Lines 13-15: Pop affordable projects (cost ≤ w), push profits to max-heap (negative for max).
• Lines 18-19: If no affordable projects, exit early.
• Line 22: Add the highest profit to capital, continue.
• Time Complexity: O(n log n + k log n)—heapify is O(n), k heap ops.
• Space Complexity: O(n)—two heaps.

It’s like a profit-maximizing engine!

Alternative Solution: Brute Force with Sorting

Why an Alternative Approach?

The brute-force solution sorts projects by profit and picks the best affordable ones each time. It’s O(n log n + k * n) time and O(n) space—simpler to grasp but slower, as it rescans the list repeatedly. Great for understanding the greedy idea without heaps.

How It Works

Imagine this as a manual search:

• Step 1: Sort projects by profit (descending).
• Step 2: For each of k picks:
• Scan for the highest-profit project you can afford.
• Mark it used, add profit to capital.
• Step 3: Repeat until k or no options.

It’s like flipping through a catalog each time!

Step-by-Step Example

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

• Init:
• Projects: [(3,1), (2,1), (1,0)] (profit, cost, index).
• Used: [False]*3.
• Pick 1:
• w = 0, can afford (1,0), w = 1, mark used.
• Pick 2:
• w = 1, can afford (2,1) or (3,1), pick (3,1), w = 4.
• Result: 4.

Code for Brute Force

Copy codeclass Solution:
    def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int:
        n = len(profits)
        projects = [(profits[i], capital[i], i) for i in range(n)]
        projects.sort(reverse=True)  # Sort by profit descending
        used = [False] * n

        for _ in range(k):
            best_profit = -1
            best_idx = -1
            # Find best affordable project
            for profit, cost, idx in projects:
                if not used[idx] and cost <= w and profit > best_profit:
                    best_profit = profit
                    best_idx = idx
            if best_idx == -1:  # No affordable project
                break
            w += best_profit
            used[best_idx] = True

        return w

• Time Complexity: O(n log n + k * n)—sorting plus k scans.
• Space Complexity: O(n)—projects and used array.

It’s a thorough but slow picker!

Comparing the Two Solutions

• Heaps (Best):
• Pros: O(n log n + k log n), efficient picks.
• Cons: Heap logic.
• Brute Force (Alternative):
• Pros: O(n log n + k * n), easy to follow.
• Cons: Slower scans.

Heaps win big!

Additional Examples and Edge Cases

• k=1, w=0, [1], [0]: 1—one project.
• k=2, w=10, [1,2], [5,5]: 13—both affordable.
• k=1, w=0, [1], [1]: 0—can’t afford.

Heaps handle them smoothly.

Complexity Recap

• Heaps: Time O(n log n + k log n), Space O(n).
• Brute Force: Time O(n log n + k * n), Space O(n).

Heaps are the efficiency kings!

Key Takeaways

• Heaps: Greedy + fast—learn them at Python Basics!
• Greedy: Pick the best now.
• Constraints: Capital drives choices.
• Fun: Investing feels good!

Final Thoughts: Grow Your Capital!

LeetCode 502: IPO in Python is a thrilling greedy challenge. Heaps make it fast and fun, while brute force keeps it simple. Want more? Try LeetCode 455: Assign Cookies or LeetCode 767: Reorganize String. Ready to invest? Head to Solve LeetCode 502 on LeetCode and maximize that capital today!