What is the time complexity of Implement Router?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Implement Router?

This problem uses Array, Hash Table, Binary Search, Design, Queue, Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Implement Router?

Insight 1: Use a queue (collections.deque in Python, deque in C++, circular array in C) to maintain the FIFO order of packets for forwarding. Insight 2: Use a hash set (set in Python and C++, custom hash table in C) to quickly check for duplicate packets based on source, destination and timestamp. This avoids linear search.

Which companies ask LeetCode 3508. Implement Router in interviews?

This problem is frequently asked at Cisco.

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Implement Router - LeetCode 3508 Solution

Implement Router - Complete Solution Guide

Implement Router is LeetCode problem 3508, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design a data structure that can efficiently manage data packets in a network router. Each data packet consists of the following attributes: source : A unique identifier for the machine that generated the packet. destination : A unique identifier for the target machine. timestamp : The time at which the packet arrived at the router. Implement the Router class: Router(int memoryLimit) : Initializes the Router object with a fixed memory limit. memoryLimit is the maximum number of packets the route

Detailed Explanation

The problem asks us to implement a `Router` class that simulates a network router managing data packets. Each packet has a `source`, `destination`, and `timestamp`. The router has a fixed `memoryLimit` on the number of packets it can store. The `Router` class needs to support four operations: `addPacket` (adds a new packet, returning true if added successfully and false if it's a duplicate), `forwardPacket` (removes and returns the oldest packet), `getCount` (returns the number of packets with a specific destination and timestamp within a range), and the constructor `Router(memoryLimit)`. `addPacket` operations arrive in increasing order of timestamp. When the router's memory is full, the oldest packet should be removed before adding a new one.

Solution Approach

The solution uses a combination of a queue, a hash set, and a hash map to maintain the packets and their timestamps. A deque (double-ended queue) is used to store the packets in FIFO order, which is important for the `forwardPacket` operation. A set is used to quickly check for duplicates when adding a new packet, providing O(1) lookup time. A dictionary (hash map) is used to maintain information about each destination, namely the timestamps and a start index. The start index optimization prevents rescanning of removed packet indices, by saving the 'first index' of valid timstamps in the `getCount` function.

Step-by-Step Algorithm

Step 1: Initialize the Router with a given `memoryLimit`, a deque (`packets_queue`), a set (`packets_set`), and a dictionary (`dest_info`).
Step 2: In `addPacket`, check if the packet already exists in `packets_set`. If it does, return `false`.
Step 3: If the router is at its `memoryLimit`, remove the oldest packet from `packets_queue` and `packets_set`. Decrement `start_idx` for the desitnation of removed packet.
Step 4: Add the new packet to `packets_queue` and `packets_set`. Also, append the packet's timestamp to the list of timestamps associated with the packet's destination in `dest_info`.
Step 5: In `forwardPacket`, if `packets_queue` is empty, return an empty list. Otherwise, remove the oldest packet from the queue and set, increment `start_idx` for the desitnation of removed packet and return its attributes as a list.
Step 6: In `getCount`, if the destination is not in `dest_info`, return 0. Otherwise, retrieve the timestamps list and the starting index associated with the destination. Perform binary search to find the indices of the first timestamp greater than or equal to `startTime` and the first timestamp greater than `endTime`. The difference between these indices gives the count of packets within the specified range.
Step 7: The C implementation uses a circular array as the queue and separate hash tables for the packets_set and dest_info. The dest_info also tracks timestamp arrays for each destination.

Key Insights

Insight 1: Use a queue (collections.deque in Python, deque in C++, circular array in C) to maintain the FIFO order of packets for forwarding.
Insight 2: Use a hash set (set in Python and C++, custom hash table in C) to quickly check for duplicate packets based on source, destination and timestamp. This avoids linear search.
Insight 3: For efficient `getCount` operation, maintain sorted lists of timestamps for each destination. This allows the use of binary search to count packets within the specified timestamp range.
Insight 4: To avoid recalculating the index where valid timestamps start for a destination, maintain an offset (`start_idx`) to mark the beginning of valid timestamp entries in the sorted timestamp list for a particular destination.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, Binary Search, Design, Queue, Ordered Set.

Companies

Asked at: Cisco.

Implement Router | LeetCode 3508

Medium

Array Hash Table Binary Search Design Queue Ordered Set

Time: O(1)Space: O(m)

Solution Code

Available in:

class Router {

    public Router(int memoryLimit) {
        
    }
    
    public boolean addPacket(int source, int destination, int timestamp) {
        
    }
    
    public int[] forwardPacket() {
        
    }
    
    public int getCount(int destination, int startTime, int endTime) {
        
    }
}

/**
 * Your Router object will be instantiated and called as such:
 * Router obj = new Router(memoryLimit);
 * boolean param_1 = obj.addPacket(source,destination,timestamp);
 * int[] param_2 = obj.forwardPacket();
 * int param_3 = obj.getCount(destination,startTime,endTime);
 */

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Problem Info

Description

Problem 3508: Implement Router - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Design Queue Ordered Set

Companies

Cisco

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3508

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

7 steps

1
Step 1: Initialize the Router with a given `memoryLimit`, a deque (`packets_queue`), a set (`packets_set`), and a dictionary (`dest_info`).
2
Step 2: In `addPacket`, check if the packet already exists in `packets_set`. If it does, return `false`.
3
Step 3: If the router is at its `memoryLimit`, remove the oldest packet from `packets_queue` and `packets_set`. Decrement `start_idx` for the desitnation of removed packet.
4
Step 4: Add the new packet to `packets_queue` and `packets_set`. Also, append the packet's timestamp to the list of timestamps associated with the packet's destination in `dest_info`.
5
Step 5: In `forwardPacket`, if `packets_queue` is empty, return an empty list. Otherwise, remove the oldest packet from the queue and set, increment `start_idx` for the desitnation of removed packet and return its attributes as a list.
6
Step 6: In `getCount`, if the destination is not in `dest_info`, return 0. Otherwise, retrieve the timestamps list and the starting index associated with the destination. Perform binary search to find the indices of the first timestamp greater than or equal to `startTime` and the first timestamp greater than `endTime`. The difference between these indices gives the count of packets within the specified range.
7
Step 7: The C implementation uses a circular array as the queue and separate hash tables for the packets_set and dest_info. The dest_info also tracks timestamp arrays for each destination.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(m)

📝

Why?

`addPacket` involves checking for duplicates in the hash set (O(1) on average) and adding to the queue and set (O(1)). Removing the oldest packet is also O(1) because we're using a deque/circular array. `forwardPacket` is O(1) because it involves removing the first element from the queue and set. `getCount` uses binary search on the timestamps associated with the destination, resulting in O(log n) time complexity, where n is the number of packets with the given destination. The space complexity is O(m) because we store, at most, `memoryLimit` packets.

📖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 list instead of a set for duplicate detection

Instead of using a set to store all the packets, one could use a list. When adding a new packet, the list would be searched to check if the packet already exists.

Time: O(n) for addPacket, O(1) for forwardPacket, and O(log n) for getCount, where n is the number of packets currently stored.Space: O(m), where m is the memoryLimit.

Trade-offs: This would reduce the space complexity slightly (no need for a set), but would significantly increase the time complexity of `addPacket` (O(n) instead of O(1)). It's a bad tradeoff if the number of addPacket operations is high.

Maintaining a single sorted timestamp list for all packets

Instead of having separate timestamp lists per destination, one could maintain a single sorted timestamp list for all packets along with a mapping from timestamp to (source, destination) pairs. Then, getCount would involve finding timestamps in the range and filtering based on destination.

Time: O(log m + k) for getCount, where m is memoryLimit and k is the number of matching timestamps after binary search.Space: O(m)

Trade-offs: The time complexity of getCount increases to O(log m + k), but simplifies the data structures. However, if the `getCount` operation is frequently called and 'k' is large, this could be less efficient than the main solution.

✨

Key Insights

1Insight 1: Use a queue (collections.deque in Python, deque in C++, circular array in C) to maintain the FIFO order of packets for forwarding.
2Insight 2: Use a hash set (set in Python and C++, custom hash table in C) to quickly check for duplicate packets based on source, destination and timestamp. This avoids linear search.
3Insight 3: For efficient `getCount` operation, maintain sorted lists of timestamps for each destination. This allows the use of binary search to count packets within the specified timestamp range.
4Insight 4: To avoid recalculating the index where valid timestamps start for a destination, maintain an offset (`start_idx`) to mark the beginning of valid timestamp entries in the sorted timestamp list for a particular destination.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: The router is empty when `forwardPacket` is called. The code handles this by returning an empty list.
Edge case 2: The destination does not exist in `dest_info` when `getCount` is called. The code handles this by returning 0.
Edge case 3: A duplicate packet is added. The code returns false without adding the packet.
Edge case 4: The `startTime` and `endTime` are such that no timestamps fall within the range. The binary search in `getCount` correctly handles this case, returning 0.
Edge case 5: The packets queue reaches the memoryLimit. The oldest packets are removed to allow new packets to be added.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to remove the oldest packet when the router is full. This can lead to incorrect counts and potential memory overflow.

✗Common mistake 2: Not handling duplicate packets correctly, leading to incorrect packet counts and potential errors in forwarding packets.

✗Common mistake 3: Incorrectly implementing the binary search in `getCount`, leading to inaccurate results. The `startTime` should be inclusive while `endTime` needs careful range consideration.

✗Common mistake 4: Failing to correctly update or track the start_idx for destination timestamps after packets are forwarded or evicted.

✗Common mistake 5: In the C implementation, not handling hash collisions, potentially leading to incorrect duplicate detection and packet management. Also, memory management is complex and prone to errors.

✅

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: Network routers use similar techniques to manage data packets, prioritize traffic, and ensure reliable data delivery. This problem is a simplified model of real-world network routing algorithms.

🏦Real-world scenario 2: Rate limiting in APIs. When handling requests, servers often limit the number of requests from a particular source within a certain time window. This problem simulates tracking and limiting those requests based on timestamps.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 28, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Implement Router"?

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

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