Youtube Design

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Find Loop in a Linked List

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 Problem Statement In this problem, we are given a linked list with a loop. We have to find the first node of the loop in the linked list. Problem Statement Understanding Let’s try to understand the problem statement with the help of examples. Suppose the given list is: According to the problem statement, we need to find the starting node of the loop. From the linked list, we can see that there is a loop in the linked list starting at node with value 3 and containing 3 nodes 3, 5, and 7. The last node of the loop points back to the first node of the loop. As node with value 3 is the first node of the loop, so we will return 3 as output. Approach and Algorithm (Floyd’s Cycle Detection) Our approach will be simple: 1) Firstly, we have to detect the loop in the given linked list. For detecting a loop in any linked list, we know the most efficient algorithm is the  Floyd Cycle detection Algorithm . 2) In Floyd's cycle detection algorithm, we initialize 2 pointers,  slow ...
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Optional Vs Null Check

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The purpose of Optional is not to replace every single null reference in your codebase but rather to help design better APIs in which—just by reading the signature of a method—users can tell whether to expect an optional value. In addition, Optional forces you to actively unwrap an Optional to deal with the absence of a value; as a result, you protect your code against unintended null pointer exceptions. String version = computer.getSoundcard().getUSB().getVersion(); This code can lead to NullPointerException. Handle NullPointerException - This code looks ugly. String version = "UNKNOWN"; if(computer != null){   Soundcard soundcard = computer.getSoundcard();   if(soundcard != null){     USB usb = soundcard.getUSB();     if(usb != null){       version = usb.getVersion();     }   } } String name = computer.flatMap(Computer::getSoundcard)                           .f...
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System Design URL Shortening Service

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URL shortening services like  bit.ly  or  TinyURL  are very popular to generate shorter aliases for long URLs. You need to design this kind of web service where if a user gives a long URL then the service returns a short URL and if the user gives a short URL then it returns the original long URL.  For example, shortening the given URL through TinyURL:  https://www.interviewready.org/get-your-dream-job-with /?ref=leftbar-rightbar We get the result given -  https://tinyurl.com/y7vg2xjl Requirement Before you jump into the solution always clarify all the assumptions you’re making at the beginning of the interview. Ask questions to identify the scope of the system. This will clear the initial doubt, and you will get to know what are the specific detail interviewer wants to consider in this service.  Given a long URL, the service should generate a shorter and unique alias of it. When the user hits a short link, the service should redirect to the origin...
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