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Elevator Requests IV - Solution & Explanation

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Problem Statement

You are given a string s consisting of lowercase English letters.

Construct a binary string by replacing each character in s with the 8-bit binary representation of its ASCII value, including leading zeros, while preserving the original order of the characters.

Return true if the resulting binary string is a palindrome. Otherwise, return false.

 

Example 1:

Input: s = "ff"

Output: true

Explanation:

  • The ASCII value of f is 102, whose 8-bit binary representation is 01100110.
  • Thus, the binary string is 0110011001100110.
  • Since this binary string is a palindrome, the output is true.

Example 2:

Input: s = "leet"

Output: false

Explanation:

  • The ASCII values of l, e, e, and t are 108, 101, 101, and 116, respectively.
  • Their 8-bit binary representations are 01101100, 01100101, 01100101, and 01110100.
  • Thus, the binary string is 01101100011001010110010101110100.
  • Since this binary string is not a palindrome, the output is false.

 

Constraints:

  • 1 <= s.length <= 100
  • s consists of lowercase English letters.

Approach Overview

Problem Overview: You're given a set of elevator requests, each with a start floor, destination floor, and time, and you need to schedule a set of elevators to minimize the total waiting time or completion time. The optimal solution must efficiently assign requests to elevators while respecting their movement constraints.

Approach 1: Brute Force (O(n^2) time, O(1) space)

Simulate every possible assignment of requests to elevators. For each request, try all elevators and check which one finishes earliest based on current position and direction. This works for tiny inputs but explodes as n grows because you iterate through all elevators for every request. It's a good starting point to verify correctness but never passes large test cases.

Approach 2: Greedy with Sorting (O(n log n) time, O(n) space)

Sort requests by arrival time and process them in order. Maintain each elevator's current floor and next available time. For each request, pick the elevator that can reach the start floor earliest (considering travel time from its current position). This is a classic greedy strategy for interval scheduling and often passes most test cases. The key insight is that you only need to compare the earliest available elevator—no need to try all combinations. However, it fails when future requests could be better served by waiting for a closer elevator, so it's not optimal in all cases.

Approach 3: Optimal with Priority Queue (O(n log n) time, O(n) space)

Use a min-heap to track each elevator's ready time and the request they're handling. Instead of scanning all elevators, you pop the elevator with the earliest available time and assign the current request to it, then push it back with its new availability. This gives a globally optimal schedule when the cost function is additive and each request is independent. The trick is to treat the elevator as a resource that becomes free after serving a request, and the heap keeps the next free elevator at the top. This approach is what interviewers expect—it shows you understand priority queues and greedy scheduling.

Recommended for interviews: The optimal priority queue solution is the one to present. Brute force shows you can reason about the problem structure, but the heap-based method demonstrates you can apply classic algorithms (greedy + priority queue) to real-world constraints. Start with brute force to confirm understanding, then jump to the optimal solution. Mention that the greedy solution fails for certain corner cases where a longer wait for a closer elevator beats picking the first free one, but the heap approach handles that by always choosing the earliest available elevator.

Relevant topics to strengthen your understanding: greedy algorithms, priority queues, and scheduling.

Solution

Following the problem statement, we replace each character of s with the 8-bit binary representation of its ASCII value (including leading zeros), concatenate them in order to obtain a binary string t, and then check whether t is a palindrome.

The time complexity is O(n) and the space complexity is O(n), where n is the length of s.

Code

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Python

Java

C++

Go

TypeScript

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute ForceO(n²)O(1)Small inputs (n ≤ 100), edge case testing
Greedy (Sort + Earliest)O(n log n)O(n)When requests are independent and elevators identical
Priority Queue (Optimal)O(n log n)O(n)General case, handles overlapping requests, best for interviews

Video Solution

Leetcode : 4030. Check ASCII Palindromic ( Weekly Contest 516 ) • Web Fusion with Bharadwaj • 44 views views

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Frequently Asked Questions

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 <code>heapq</code> module; in Java, use <code>PriorityQueue</code>; in C++, use <code>priority_queue</code>. 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).

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