Is Elevator Requests I easy or hard?

Elevator Requests I is rated as an Easy problem with an 83.5% acceptance rate. It is a straightforward simulation task that requires understanding of basic iteration and conditional logic. The optimal solution is intuitive and can be implemented in a few lines of code, making it a good warm-up for more complex problems.

Elevator Requests I Python/Java solution

In Python, you can implement the single pass simulation with a simple for loop. In Java, use a for-each loop over the requests array. The logic is identical: track current floor and count, compute distances, and update accordingly. Both solutions run in O(n) time and O(1) space.

How to solve Elevator Requests I in O(n)?

To solve in O(n), maintain a variable for the current floor and a counter for served requests. For each request, calculate the absolute difference between the current floor and the request's start floor, then add the distance from start to end. If the total distance is within the allowed time, update the current floor to the end floor and increment the counter. This single pass avoids sorting and nested loops.

What is the best approach for Elevator Requests I?

The best approach is the single pass simulation, which runs in O(n) time and O(1) space. It iterates through each request once, updating the elevator's current floor and counting served requests based on whether the travel distance fits within the total time. This is optimal because it avoids unnecessary sorting or nested loops.

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

Elevator Requests I is a general simulation problem that tests basic algorithmic thinking. While not specifically tagged for these companies, similar simulation problems are common in technical interviews at top tech companies. The optimal O(n) approach demonstrates the kind of clean, efficient coding skills interviewers look for.

What data structure is used in Elevator Requests I?

The optimal solution uses no complex data structures—just basic variables to track the current floor and the count of served requests. This makes the solution extremely memory-efficient with O(1) space. Some alternative approaches might use sorting, which requires an array or list, but the single pass simulation avoids that.

What is the time complexity of Elevator Requests I?

The optimal solution has a time complexity of O(n), where n is the number of requests. This is achieved by a single pass simulation that processes each request in constant time. The space complexity is O(1) as no additional data structures are used.

Description Hints 2 Solution Video 18:51

4020. Elevator Requests I

Easy84.1% acceptance

You are given an integer n denoting the number of floors in a building, where the floors are numbered from 0 to n - 1.

You are also given an integer array requests, where requests represents the sequence of floor requests.

An elevator starts at floor 0, and follows these rules:

The elevator moves one floor per second.
The elevator serves requests in the given order.
If the elevator is already on the requested floor, no movement is needed.
After serving a request, the elevator immediately starts moving toward the next request.

Return the total time (in seconds) required to serve all requests.

Example 1:

Input: n = 5, requests = [2,1,4,3]

Output: 7

Explanation:

requests[0] = 2: Moving from floor 0 to floor 2 takes 2 seconds.
requests[1] = 1: Moving from floor 2 to floor 1 takes 1 second.
requests[2] = 4: Moving from floor 1 to floor 4 takes 3 seconds.
requests[3] = 3: Moving from floor 4 to floor 3 takes 1 second.

The total time required is 2 + 1 + 3 + 1 = 7 seconds.

Example 2:

Input: n = 3, requests = [2,0,0]

Output: 4

Explanation:

requests[0] = 2: Moving from floor 0 to floor 2 takes 2 seconds.
requests[1] = 0: Moving from floor 2 to floor 0 takes 2 seconds.
requests[2] = 0: No movement is needed.

The total time required is 2 + 2 + 0 = 4 seconds.

Constraints:

1 <= n <= 100
1 <= requests.length <= 100
0 <= requests[i] <= n - 1

Topics

Array Simulation

Hints open one at a time

Hint 1

Keep track of the elevator's current floor, starting from floor 0.

Hint 2

For each requested floor, add the absolute difference between it and the current floor to the answer, then update the current floor.

18:51

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def elevatorRequests(self, n: int, requests: list[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

5
[2,1,4,3]

Input

3
[2,0,0]

Your code is saved on this device.Solution