How do you solve Find Servers That Handled Most Number of Requests on LeetCode?

Find Servers That Handled Most Number of Requests (LeetCode #1606) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: In this approach, we simply iterate through each request and check each server to see if it can handle the request. If a server is busy, we check the next server until we find one that is available or exhaust all options.

What is the time complexity of Find Servers That Handled Most Number of Requests?

The optimal solution for Find Servers That Handled Most Number of Requests runs in O(n²) time and uses O(1) space. This complexity arises because for each request, we may need to check all servers in the worst case, leading to a quadratic time complexity.

What algorithm or data structure is used to solve Find Servers That Handled Most Number of Requests?

Find Servers That Handled Most Number of Requests uses the following concepts: Array, Heap (Priority Queue), Simulation, Ordered Set. The recommended patterns to study are: Heap, Greedy.

What are the interview tips for Find Servers That Handled Most Number of Requests?

Explain your thought process clearly while coding. Discuss edge cases, such as when all servers are busy. Practice similar problems to get comfortable with priority queues.

What are common mistakes when solving Find Servers That Handled Most Number of Requests?

Not managing server availability correctly, leading to incorrect counts. Failing to account for dropped requests.

Find Servers That Handled Most Number of Requests — LeetCode #1606 (Hard)

Tags: Array, Heap (Priority Queue), Simulation, Ordered Set

Related patterns: Heap, Greedy

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In this approach, we simply iterate through each request and check each server to see if it can handle the request. If a server is busy, we check the next server until we find one that is available or exhaust all options.

This complexity arises because for each request, we may need to check all servers in the worst case, leading to a quadratic time complexity.

Step 1: Initialize an array to keep track of the end times for each server.
Step 2: For each request, check if the (i % k) th server is available. If not, check the next servers in a circular manner.
Step 3: If a server is found to be available, update its end time based on the current request's load and increment the count of handled requests for that server.

1. Request 0 arrives at time 1, assigned to server 0 (endTimes: [6, 0, 0], count: [1, 0, 0]) 2. Request 1 arrives at time 2, assigned to server 1 (endTimes: [6, 4, 0], count: [1, 1, 0]) 3. Request 2 arrives at time 3, assigned to server 2 (endTimes: [6, 4, 6], count: [1, 1, 1]) 4. Request 3 arrives at time 4, server 0 is busy, assigned to server 1 (endTimes: [6, 4, 6], count: [1, 2, 1]) 5. Request 4 arrives at time 5, server 0 is busy, assigned to server 2 (endTimes: [6, 4, 8], count: [1, 2, 2]) 6. Final counts: [1, 2, 2], busiest server is 1.

Common Mistakes

Not managing server availability correctly, leading to incorrect counts.
Failing to account for dropped requests.

Interview Tips

Explain your thought process clearly while coding.
Discuss edge cases, such as when all servers are busy.
Practice similar problems to get comfortable with priority queues.

#1606

Find Servers That Handled Most Number of Requests

Hard

Array↗Heap (Priority Queue)↗Simulation↗Ordered Set↗Heap↗Greedy↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

3 steps

1Step 1: Initialize an array to keep track of the end times for each server.
2Step 2: For each request, check if the (i % k) th server is available. If not, check the next servers in a circular manner.
3Step 3: If a server is found to be available, update its end time based on the current request's load and increment the count of handled requests for that server.

solution.py11 lines

1def busiestServers(k, arrival, load):
2    end_times = [0] * k
3    count = [0] * k
4    for i in range(len(arrival)):
5        while i < len(arrival) and end_times[i % k] > arrival[i]:
6            i += 1
7        if i < len(arrival):
8            end_times[i % k] = arrival[i] + load[i]
9            count[i % k] += 1
10    max_requests = max(count)
11    return [i for i in range(k) if count[i] == max_requests]

ℹ

Complexity note: This complexity arises because for each request, we may need to check all servers in the worst case, leading to a quadratic time complexity.

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