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

The optimal solution has a time complexity of O(n log k).

What data structures are used to solve Find Servers That Handled Most Number of Requests?

This problem uses Array, Greedy, Heap (Priority Queue), Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Servers That Handled Most Number of Requests?

Insight 1: The key is to efficiently track the availability of servers and their completion times. This requires using appropriate data structures like priority queues (min-heaps). Insight 2: The round-robin assignment strategy and wrap-around behavior can be simulated efficiently using the modulo operator (%) and by maintaining two separate sorted data structures for available servers based on server ID relative to `i % k`.

Which companies ask LeetCode 1606. Find Servers That Handled Most Number of Requests in interviews?

This problem is frequently asked at Capital One, Cisco, Citadel, Visa, Wish.

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Find Servers That Handled Most Number of Requests - LeetCode 1606 Solution

Find Servers That Handled Most Number of Requests - Complete Solution Guide

Find Servers That Handled Most Number of Requests is LeetCode problem 1606, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have k servers numbered from 0 to k-1 that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time . The requests are assigned to servers according to a specific algorithm: The i th (0-indexed) request arrives. If all servers are busy, the request is dropped (not handled at all). If the (i % k) th server is available, assign the request to that server. Otherwise, assign the request to the nex

Detailed Explanation

The problem asks us to simulate a system of 'k' servers handling incoming requests. Each request has an arrival time and a load (processing time). The requests are assigned to servers in a round-robin fashion (i % k). If the target server is busy, we try the next available server, wrapping around to the beginning if necessary. If all servers are busy, the request is dropped. The goal is to find the server(s) that handled the most requests and return their IDs.

Solution Approach

The provided solutions use a combination of priority queues (min-heaps) and arrays to simulate the server allocation and request handling process. A priority queue `busy_servers` stores servers that are currently processing requests, sorted by their free time. To track available servers and handle the round-robin allocation, two priority queues, `available_ge` and `available_lt`, are maintained. The `available_ge` heap stores available server IDs that are greater than or equal to the target server ID (`i % k`), and `available_lt` stores those less than `i % k`. When a request arrives, servers are freed up if their free time is earlier than the current arrival time. An available server is retrieved from `available_ge` or `available_lt`. The number of requests handled by each server is tracked using an array. Finally, the servers with the maximum number of requests are identified and returned.

Step-by-Step Algorithm

Step 1: Initialize an array `counts` of size 'k' to store the number of requests handled by each server, initially set to zero.
Step 2: Initialize a min-heap `busy_servers` to store the free time and server ID of currently busy servers.
Step 3: Initialize two min-heaps, `available_ge` and `available_lt`, to store available server IDs. `available_ge` contains server IDs >= `i % k`, and `available_lt` contains server IDs < `i % k`. Initially, all servers are in `available_ge`.
Step 4: Iterate through the `arrival` and `load` arrays. For each request:
Step 5: Update the `available_ge` and `available_lt` heaps based on `i % k` and `(i-1) % k`. If we wrap around, merge the available heaps and re-partition.
Step 6: Free up any servers in `busy_servers` that have finished processing before the current request's arrival time and add them to the correct available heap.
Step 7: Attempt to assign the request to an available server. First, check `available_ge`, then `available_lt`. If no server is available, the request is dropped.
Step 8: If a server is available, increment its request count in `counts`, calculate its free time, and add it to `busy_servers`.
Step 9: After processing all requests, find the maximum number of requests handled by any server.
Step 10: Return a list of the server IDs that handled the maximum number of requests.

Key Insights

Insight 1: The key is to efficiently track the availability of servers and their completion times. This requires using appropriate data structures like priority queues (min-heaps).
Insight 2: The round-robin assignment strategy and wrap-around behavior can be simulated efficiently using the modulo operator (%) and by maintaining two separate sorted data structures for available servers based on server ID relative to `i % k`.
Insight 3: We need to handle the edge case where all servers are busy, resulting in dropped requests. Also, the wrap-around logic needs to be carefully implemented to avoid incorrect server assignments.

Complexity Analysis

Time Complexity: O(n log k)

Space Complexity: O(k)

Topics

This problem involves: Array, Greedy, Heap (Priority Queue), Ordered Set.

Companies

Asked at: Capital One, Cisco, Citadel, Visa, Wish.

Find Servers That Handled Most Number of Requests | LeetCode 1606

Hard

Array Greedy Heap (Priority Queue)Ordered Set

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<Integer> busiestServers(int k, int[] arrival, int[] load) {
        int[] counts = new int[k];
        PriorityQueue<int[]> busyServers = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])); // min-heap of {free_time, server_id}

        // Two priority queues for available servers, partitioned by i % k
        // available_ge for servers with id >= i % k
        // available_lt for servers with id < i % k
        PriorityQueue<Integer> availableGe = new PriorityQueue<>();
        PriorityQueue<Integer> availableLt = new PriorityQueue<>();

        for (int i = 0; i < k; i++) {
            availableGe.add(i);
        }

        for (int i = 0; i < arrival.length; i++) {
            int currentTime = arrival[i];
            int currentLoad = load[i];
            int target = i % k;

            // Adjust heaps for new target for i > 0.
            // For i=0, target=0, and the initial setup is correct.
            if (i > 0) {
                int prevTarget = (i - 1) % k;
                if (target > prevTarget) {
                    // Move servers with IDs in [prev_target, target) from ge to lt
                    while (!availableGe.isEmpty() && availableGe.peek() < target) {
                        availableLt.add(availableGe.poll());
                    }
                } else if (target < prevTarget) { // Wrap-around case
                    // Merge all available servers and re-partition
                    List<Integer> temp = new ArrayList<>();
                    while(!availableGe.isEmpty()){
                        temp.add(availableGe.poll());
                    }
                     while(!availableLt.isEmpty()){
                        temp.add(availableLt.poll());
                    }
                    Collections.sort(temp);
                    availableGe.addAll(temp);
                    availableLt.clear();
                    
                    while (!availableGe.isEmpty() && availableGe.peek() < target) {
                        availableLt.add(availableGe.poll());
                    }
                }
            }

            // Free up servers that have completed their tasks
            while (!busyServers.isEmpty() && busyServers.peek()[0] <= currentTime) {
                int[] serverInfo = busyServers.poll();
                int serverId = serverInfo[1];
                // Place the now-free server into the correct available heap
                if (serverId >= target) {
                    availableGe.add(serverId);
                } else {
                    availableLt.add(serverId);
                }
            }

            // Find an available server for the current request
            Integer serverToUse = null;
            if (!availableGe.isEmpty()) {
                serverToUse = availableGe.poll();
            } else if (!availableLt.isEmpty()) {
                serverToUse = availableLt.poll();
            }

            // If a server is found, assign the request. Otherwise, drop it.
            if (serverToUse != null) {
                counts[serverToUse] += 1;
                int freeTime = currentTime + currentLoad;
                busyServers.add(new int[]{freeTime, serverToUse});
            }
        }

        int maxRequests = 0;
        for (int count : counts) {
            maxRequests = Math.max(maxRequests, count);
        }

        List<Integer> busiest = new ArrayList<>();
        for (int j = 0; j < k; j++) {
            if (counts[j] == maxRequests) {
                busiest.add(j);
            }
        }

        return busiest;
    }
}

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Problem Info

Description

Problem 1606: Find Servers That Handled Most Number of Requests - Detailed explanation and solution approaches.

Topics

Array Greedy Heap (Priority Queue)Ordered Set

Companies

Capital OneCiscoCitadelVisaWish

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:1606

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize an array `counts` of size 'k' to store the number of requests handled by each server, initially set to zero.
2
Step 2: Initialize a min-heap `busy_servers` to store the free time and server ID of currently busy servers.
3
Step 3: Initialize two min-heaps, `available_ge` and `available_lt`, to store available server IDs. `available_ge` contains server IDs >= `i % k`, and `available_lt` contains server IDs < `i % k`. Initially, all servers are in `available_ge`.
4
Step 4: Iterate through the `arrival` and `load` arrays. For each request:
5
Step 5: Update the `available_ge` and `available_lt` heaps based on `i % k` and `(i-1) % k`. If we wrap around, merge the available heaps and re-partition.
6
Step 6: Free up any servers in `busy_servers` that have finished processing before the current request's arrival time and add them to the correct available heap.
7
Step 7: Attempt to assign the request to an available server. First, check `available_ge`, then `available_lt`. If no server is available, the request is dropped.
8
Step 8: If a server is available, increment its request count in `counts`, calculate its free time, and add it to `busy_servers`.
9
Step 9: After processing all requests, find the maximum number of requests handled by any server.
10
Step 10: Return a list of the server IDs that handled the maximum number of requests.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log k)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by the heap operations performed for each request. Each heap insertion/deletion/peek takes O(log k) time, and these operations are performed within the main loop that iterates 'n' times. The space complexity is determined by the size of the data structures used to store the server information. The `counts` array and the heaps (`busy_servers`, `available_ge`, `available_lt`) each consume space proportional to 'k'.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Balanced Tree (e.g., TreeSet in Java)

Instead of two priority queues (`available_ge` and `available_lt`), a single self-balancing binary search tree (like TreeSet) could be used to store the available servers. We can efficiently find the smallest available server ID >= i % k using methods like `ceiling` (Java) or `lower_bound` (C++).

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

Trade-offs: The time complexity remains the same, but the implementation might be slightly simpler than using two separate heaps for managing available servers. It simplifies the wrap around handling as well, although, can be slower due to the added complexity of the tree operations.

✨

Key Insights

1Insight 1: The key is to efficiently track the availability of servers and their completion times. This requires using appropriate data structures like priority queues (min-heaps).
2Insight 2: The round-robin assignment strategy and wrap-around behavior can be simulated efficiently using the modulo operator (%) and by maintaining two separate sorted data structures for available servers based on server ID relative to `i % k`.
3Insight 3: We need to handle the edge case where all servers are busy, resulting in dropped requests. Also, the wrap-around logic needs to be carefully implemented to avoid incorrect server assignments.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: All servers are busy. In this case, the algorithm correctly drops the request because it cannot find an available server in either `available_ge` or `available_lt`.
Edge case 2: The number of requests is much smaller than the number of servers (n << k). The algorithm still performs correctly, although the heap operations might be less efficient because the heaps might not be fully populated.
Edge case 3: Some servers handle zero requests. The algorithm correctly identifies the busiest servers even if some have zero requests since it considers the maximum count.
Edge case 4: All servers handle the same number of requests. The algorithm correctly returns a list containing all server IDs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly handling the wrap-around logic when assigning requests to servers. This can lead to requests being assigned to the wrong servers or dropped unnecessarily.

✗Common mistake 2: Failing to free up servers that have completed their tasks before attempting to assign new requests. This can lead to all servers being incorrectly marked as busy and requests being dropped when available servers exist.

✗Common mistake 3: Incorrectly implementing the heap operations or comparing the free times of busy servers, leading to incorrect server scheduling.

✗Common mistake 4: Not considering the case where multiple servers handle the same maximum number of requests and not returning all of their IDs.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Load balancing in a distributed system. The servers can be thought of as machines in a data center, and the requests can be tasks that need to be processed. This algorithm can help distribute the load evenly across the servers to maximize throughput and minimize response time.

🏦Real-world scenario 2: Task scheduling in an operating system. The servers can be thought of as CPU cores, and the requests can be processes that need to be executed. The algorithm can help schedule processes to CPU cores in a way that minimizes idle time and maximizes CPU utilization.

🎮Real-world scenario 3: Handling customer support tickets. Servers are support agents and requests are incoming support requests. The algorithm distributes requests in a round robin manner but tries to keep agents busy to minimize wait times.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find Servers That Handled Most Number of Requests"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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