How do you solve IPO on LeetCode?

IPO (LeetCode #502) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + k log k) time complexity and O(n) space complexity. Key insight: Maximizing profit requires careful selection of projects based on current capital.

What is the brute force solution for IPO?

The brute force approach for IPO is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible combinations of projects to see which combination yields the maximum capital. This is straightforward but inefficient as it explores every possible way to select projects. The optimal Optimal Solution approach improves this to O(n log n + k log k).

What algorithm or data structure is used to solve IPO?

IPO uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithm, Heap (Priority Queue).

What are the interview tips for IPO?

Always clarify the constraints and requirements before jumping into coding. Think about edge cases, such as when k is larger than the number of projects. Explain your thought process clearly while coding, as it helps the interviewer understand your approach.

What are common mistakes when solving IPO?

Not considering the capital constraints when selecting projects. Failing to use a max-heap or priority queue to efficiently manage project profits.

#502

IPO

Hard

ArrayGreedySortingHeap (Priority Queue)Greedy AlgorithmHeap (Priority Queue)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + k log k)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + k log k)Space O(n)

The optimal solution uses a greedy approach with a max-heap to always select the most profitable projects that can be started with the current capital. This ensures we maximize our capital efficiently.

⚙️

Algorithm

3 steps

1Step 1: Pair the profits with their respective capital requirements and sort these pairs by capital.
2Step 2: Use a max-heap to keep track of the profits of projects that can be started with the current capital.
3Step 3: For up to k projects, extract the maximum profit from the heap and add it to the current capital.

solution.py14 lines

1# Full working Python code
2import heapq
3
4def maxCapitalOptimal(k, w, profits, capital):
5    projects = sorted(zip(capital, profits))
6    max_heap = []
7    index = 0
8    for _ in range(k):
9        while index < len(projects) and projects[index][0] <= w:
10            heapq.heappush(max_heap, -projects[index][1])  # Push negative profit for max-heap
11            index += 1
12        if max_heap:
13            w -= heapq.heappop(max_heap)
14    return w

ℹ

Complexity note: The time complexity is O(n log n) for sorting the projects and O(k log k) for managing the max-heap, making it efficient for larger inputs.

1Maximizing profit requires careful selection of projects based on current capital.
2Using a greedy approach with a max-heap allows for efficient selection of the most profitable projects.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.