IPO

You are given n projects. Project i produces a pure profit of profits[i] and requires at least capital[i] current capital before it can be started.
You begin with capital w. A project's capital requirement is only an eligibility threshold; it is not spent. After completing a project, immediately add its profit to your capital. Each project may be completed at most once.
Choose at most k distinct projects to maximize your final capital. Return that maximum final capital.

Function
findMaximizedCapital(k: int, w: int, profits: int[], capital: int[]) → int

Examples
Example 1
k = 2
w = 0
profits = [1,2,3]
capital = [0,1,1]
return = 4
Complete project 0 to grow capital from 0 to 1. Projects 1 and 2 then become affordable; choosing project 2 produces final capital 4.
Example 2
k = 3
w = 0
profits = [1,2,3]
capital = [1,1,2]
return = 0
No project is affordable with initial capital 0, so the final capital remains 0.

Constraints
1 <= profits.length = capital.length <= 10^5
1 <= k <= profits.length
0 <= w <= 10^9
0 <= profits[i] <= 10^9
0 <= capital[i] <= 10^9
The maximum final capital fits in a signed 32-bit integer.

The trick: at every step, among projects you can afford, take the one with the biggest profit. Profit only grows your capital, so greedy is safe. Sort projects by capital requirement. Keep a pointer into that sorted list and a max-heap of profits. Each of up to k rounds, push every project whose capital is now <= w onto the heap, then pop the largest profit and add it to w. If the heap is empty, stop early, since nothing else unlocks. That's O(n log n) total. Common pitfalls: subtracting capital as if it were spent (it's only a threshold), forgetting to break when the heap is empty, and using a min-heap by mistake. Example 2 is the edge case: nothing is affordable at w = 0, so you return 0. If your mind goes blank on the live OA, StealthCoder can surface this sort plus heap structure quickly.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.