Is Elevator Requests IV easy or hard?

It's marked as Hard on FleetCode (acceptance ~74%). The brute force is trivial, but the optimal O(n log n) solution requires understanding of greedy scheduling and heap operations, making it a solid challenge for coding interviews.

Elevator Requests IV Python/Java solution

The priority queue solution is language-agnostic. In Python, use the heapq module; in Java, use PriorityQueue ; in C++, use priority_queue . The logic remains the same: sort requests by arrival, keep a heap of elevator availability times, and assign each request to the earliest free elevator.

How to solve Elevator Requests IV in O(n log n)?

Use a min-heap (priority queue) to track elevator availability times. Sort requests by arrival time, then for each request, pop the elevator that becomes free earliest (or the one that can reach the start floor soonest). Assign the request, update the elevator's availability, and push back into the heap. This yields O(n log n) time and O(n) space.

Is Elevator Requests IV asked at Google/Amazon/Meta?

Yes, scheduling problems like this appear frequently at top tech companies, especially as system design and optimization questions. The use of priority queues is a common pattern tested at Google and Amazon. While this exact problem may not appear verbatim, the underlying concepts are essential for interviews.

Is Elevator Requests IV asked at Google/Meta/Amazon?

Elevator scheduling problems are common in system design interviews, but this specific LeetCode-style problem might appear as a coding question at top companies like Google and Amazon because it tests greedy algorithms and priority queue usage. The acceptance rate is high (74%), so it's expected to be solvable in an interview setting.

What is the time complexity of Elevator Requests IV?

The optimal solution runs in O(n log n) time, where n is the number of requests. The space complexity is O(n) for the priority queue and sorted request list. Brute force would be O(n²) time and O(1) space.

How to solve Elevator Requests IV in O(n log n) time?

Sort all requests by arrival time and maintain a min-heap of elevator ready times. For each request, pop the elevator with the earliest ready time, assign the request, compute its new ready time (ready + travel distance + stop time), and push back. This runs in O(n log n) because each heap operation is O(log k) and there are n requests.

What data structures are used in Elevator Requests IV?

The optimal solution uses a min-heap (priority queue) to efficiently fetch the elevator that becomes free earliest. Sorting the requests by arrival time is also key. This combination reduces the O(n²) brute force to O(n log n).

What is the optimal approach for Elevator Requests IV?

The optimal approach uses a min-heap (priority queue) to track the next available elevator. By sorting requests by arrival time and always assigning the current request to the elevator that becomes free earliest, you achieve O(n log n) time and O(n) space, which is optimal for this scheduling problem.

What data structure is used in Elevator Requests IV optimal solution?

The optimal solution uses a min-heap (priority queue) to manage elevator availability. Each elevator is represented by its next free time, and the heap allows you to fetch the earliest available elevator in O(log k) time, where k is the number of elevators. This reduces overall complexity from O(n·k) to O(n log k).

4030. Elevator Requests IV - Practice Online

4030. Elevator Requests IV

Hard74.2% AcceptancePremium

PremiumFree on FleetCode

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