Transcript for transcript_1G8N7YGH9OJw-mfL7gDtLxAnj4ue-AThI_1_1936_181fab149276_6.txt

transcript_1G8N7YGH9OJw-mfL7gDtLxAnj4ue-AThI_1_1936_181fab149276_6.txt

[00:03:26]  Hello.
[00:03:27]  Good evening, sir.
[00:03:29]  Good evening, Bridges.
[00:03:32]  How are you?
[00:03:33]  I'm Pansit.
[00:03:34]  How are you?
[00:03:35]  I'm also fine.
[00:03:36]  So let me introduce myself.
[00:03:39]  I am Pankaj Jadhav and I am here in Xinex from past three months.
[00:03:44]  I am working on AI agents, building AI agents and all.
[00:03:49]  So now you can tell me about yourself.
[00:03:52]  Good morning, sir.
[00:03:54]  I am Bridges Gupta from Gorakhpur, Uttar Pradesh.
[00:03:59]  I have done my schooling from Gorakhpur itself.
[00:04:02]  Currently, I am in final year from MNAD, Allahabad.
[00:04:06]  Sir, I have good knowledge of DSA and I have been interested in Manstep.
[00:04:13]  I have worked on several Manstep projects which I have listed in my resume.
[00:04:19]  Apart from these technical skills, I am also a member of Anokepal where we give free coaching to the students near our college.
[00:04:28]  And I am also a counseling member at the Super 100 which is in Kanpur.
[00:04:34]  So that's all about me.
[00:04:36]  So you are in with sim so I was a sir a semester a semester yes okay so can you just show me your resume
[00:04:49]  like I currently I don't have yes sir I present your resume
[00:04:56]  Yes, sir.
[00:04:59]  They said this is, this is the name.
[00:05:07]  So, this slide will be up.
[00:05:10]  What?
[00:05:12]  Sir, should I slide the page up on the top or it is okay?
[00:05:17]  No, it's fine.
[00:05:17]  Like you can share your screen and show me your resume.
[00:05:20]  Okay, sir.
[00:05:44]  Okay, you can tell me about your Recycle Hub project.
[00:05:47]  Sir, in Recycle Hub, we have developed a platform where a dealer
[00:05:59]  can accept the request which will be made by any customer for the scrap pickup.
[00:06:08]  In this project, we have implemented the logic where any customer can make a request and the nearby dealer
[00:06:21]  will get notified and they will just accept the request and will go to the pickup for that request.
[00:06:29]  We have developed this project in months that we have used Google OAuth 2.0 for the mail login.
[00:06:43]  And we have also integrated a Roser payment gateway in this project for the payment which will be done from the merchant side.
[00:06:54]  Okay, so you have mentioned here about multi-role system with role-based access and feature using React Reader.
[00:07:02]  So can you please explain me how did you implement this role-based system, multi-role system with role-based access?
[00:07:11]  Yes, sir.
[00:07:11]  Sir, in this, sir, I have used reject knowledge in which I have made a, I have used a Redux toolkit
[00:07:25]  where we made a store for the specific users.
[00:07:30]  And we have used something like else and if and else logic.
[00:07:37]  So if the user is a dealer, then we have shown the dealer part of the logic.
[00:07:47]  And if the user is customer, then we have shown the customer part logic in the UI.
[00:07:56]  And similarly for the backend data, it has managed.
[00:08:02]  So suppose you are storing it in React Redux.
[00:08:06]  Yes, sir.
[00:08:06]  If you refresh a page, then whatever you have stored in the React Redux, it will go away.
[00:08:13]  Sir, when any customer will log in, then its detail will be fetched from the database.
[00:08:19]  And based upon that, the Redux...
[00:08:24]  the data of the user will be stored in the store of the Redux, Redux and accordingly the whole function will work for that user.
[00:08:35]  Okay, so you have mentioned about MongoDB.
[00:08:39]  So can you tell me something about MongoDB?
[00:08:42]  Sir, MongoDB is a NoSQL database in which it don't have any fixed schema.
[00:08:54]  We can use more than one schema type within a single database.
[00:09:00]  And sir, I have used here because of some of its features.
[00:09:05]  First of all, it is very user friendly for the beginners.
[00:09:10]  And second, it is very fit for the horizontal scaling.
[00:09:16]  That's why I have used MongoDB here for my project.
[00:09:23]  So you have told about horizontal scaling.
[00:09:25]  So can you brief me about what is horizontal scaling?
[00:09:29]  So like if the number of users get increased for any application, then
[00:09:39]  that amount of, then in a single database, we cannot accumulate all the data of all the users.
[00:09:48]  So we just divide the data of the user in more than one nodes, one DB nodes.
[00:09:57]  That is the horizontal scaling.
[00:10:00]  and can you tell me about the vertical scaling as well yes sir in vertical vertical scaling we just increase the capacity of the uh dv like we will increase the uh uh
[00:10:14]  size of the dv uh likewise we just increase the capacity of the node
[00:10:21]  Okay, so does MongoDB follows asset properties?
[00:10:27]  Yes, sir.
[00:10:27]  It follows asset properties as like SQL databases, but not much strictly as they follow.
[00:10:39]  Okay, so can you explain all the asset properties?
[00:10:44]  Yes, sir.
[00:10:46]  Atomicity, A means atomicity, means atomicity means in a single system, each process should be, each transaction should be atomic.
[00:10:57]  Means when any transaction will occur, then it should complete in a single instance.
[00:11:04]  Otherwise, its failure will corrupt the database.
[00:11:10]  C means consistency, means our database should be consistent before and after the transaction will happen.
[00:11:18]  Then isolation means in a single system more than one transition should happen and they should be unaware of each other and each will
[00:11:30]  It will occur parallelly without disturbing each other.
[00:11:35]  And D means...
[00:11:39]  Durability means when any transition will happen and suppose if any
[00:11:45]  power failure will occur or any other failure will occur then the changes which was made in the database while the transition was happening then it should be persistent in the database even after the failure of the system.
[00:12:01]  Okay.
[00:12:03]  So.
[00:12:06]  I am pasting one DSA question in the chat.
[00:12:12]  Can you solve this question?
[00:12:25]  the option for
[00:12:29]  for chat section.
[00:12:32]  Yes, I am getting.
[00:12:36]  I said given an array of integer nums and integer target, return the indices of two numbers such that they add up to target.
[00:12:47]  You imagine that each input would have exactly one solution.
[00:12:50]  You mean to do the same thing twice.
[00:12:55]  Yes, sir.
[00:12:55]  Means I have given an array of integers and I have to just find the any two indices whose sum is equal to the target value.
[00:13:07]  Yes.
[00:13:08]  Yes.
[00:13:09]  Focus.
[00:13:12]  Yes, sir.
[00:13:13]  So how will you solve it?
[00:13:20]  Sir, what I am thinking is first I will just sort the array.
[00:13:25]  Then I will take two variables, one from the beginning and other from the end.
[00:13:31]  And I will start iterating from beginning, from left to right for the first pointer and from right to left for the second pointer.
[00:13:40]  And since it is given that solution is unique, so I will just put the condition.
[00:13:50]  For moving the both the pointers and i will just accumulate both the pointer towards each other and if i will get any answer then i will just return it otherwise i will return minus one if it will not find found
[00:14:04]  okay so how will you move the left pointer and the right pointer sir uh if uh suppose i have given sir can i use my new tool
[00:14:12]  Yes, sure.
[00:14:14]  So suppose I have a of I, I is the on the left side, means starting point and J is on the ending point, I of I, A of I plus A of J.
[00:14:25]  And if this value is greater than K, means target value.
[00:14:32]  Means the J point is further.
[00:14:40]  So I have to just decrease the value.
[00:14:45]  J pointer or I will just increase the I pointer and similarly for the condition for the smaller.
[00:14:59]  Okay, understood, understood.
[00:15:02]  Also, we can also use other approach of using set to just search.
[00:15:11]  First, we will just put the elements in the set.
[00:15:14]  And again, when we will iterate over the array for the current element, if target minus current element is present in the set, then we will return that value.
[00:15:25]  But in that case, I think the answer can be wrong because can you tell me why?
[00:15:30]  Sir, in which case?
[00:15:32]  The set case that you have told.
[00:15:34]  Sir, in set case, the answer will be wrong.
[00:15:36]  Like in some test cases, it will be wrong.
[00:15:42]  Yes, sir.
[00:15:43]  It will be wrong when suppose I have 10 and I have pushed 5 and 5 is occurring only one time.
[00:15:50]  And when I will check 10 minus 5 equal to 5, it will be present in the set.
[00:15:56]  Yes.
[00:15:56]  It will give us.
[00:15:57]  So, sir, we can use option that when I will move from 0 index to till the n minus 1 index, I will when we will move, then we will just remove that element from the set.
[00:16:11]  That element and the previous elements so that we will just have only the only the element right after it so this will be this can be solved using that
[00:16:22]  Okay, so the earlier solution that you have told me, what is the time complexity of that solution?
[00:16:28]  Sir, it will be big O of n, sir.
[00:16:33]  Which solution?
[00:16:34]  Sir, that one, sir, using two-pointer.
[00:16:38]  Using two-pointer?
[00:16:39]  Yes, sir.
[00:16:41]  How?
[00:16:44]  Sir, can I draw my own solution on the paper?
[00:16:52]  Okay.
[00:16:53]  So, like you told me that you will first sort the array and then use the two pointer.
[00:16:57]  Yes, sir.
[00:16:58]  So, what will be the time complexity?
[00:17:01]  Sir, if we say big O of n, average time complexity will be big O of n.
[00:17:08]  I don't think so because you are sorting the...
[00:17:14]  Sorry, sir.
[00:17:15]  Yes, sir.
[00:17:16]  It will be sort of n log n.
[00:17:19]  Yes, yes, yes, yes.
[00:17:21]  Yes, sir.
[00:17:21]  Sorry.
[00:17:22]  I just missed it.
[00:17:24]  Okay.
[00:17:25]  And suppose...
[00:17:28]  If okay that is fine and let me
[00:17:37]  So what other projects you have done?
[00:17:38]  Can you just go to again your resume?
[00:17:42]  Yes, sir.
[00:17:43]  I have used it with the setu.
[00:17:47]  Sir, it was a collaborative project of a team in which we have just made a project where students can manage their own profile.
[00:18:00]  They can track their progress of day-to-day activities.
[00:18:05]  Or they can just schedule their
[00:18:14]  targets like for the weekly plan or the monthly plan and also we have used a feature where a professor can take the attendance of the students present
[00:18:30]  in the class.
[00:18:31]  So this we have made.
[00:18:36]  Well done, interactive, creative dashboard feature in Redline.
[00:18:40]  So how did you implement this socket.io?
[00:18:45]  Sir, actually, sir, in this, I have contributed as a front end, so I don't have much knowledge about the implementation of socket.io in this project.
[00:18:59]  Ok.
[00:19:05]  Sir, it was a hackathon in our college.
[00:19:08]  So, sir, we have participated as a team of four members.
[00:19:12]  So, sir, we have different tasks to do in the project.
[00:19:16]  Okay, understood, understood.
[00:19:19]  So, suppose...
[00:19:26]  So you know about the race condition that happens in, do you know about the race condition?
[00:19:32]  Yes, I know about the race condition.
[00:19:34]  Yes, can you explain that?
[00:19:37]  Sir, in the multi-thread system, suppose if there are more than two threads
[00:19:45]  who are trying to acquire the same section of the code, same piece of the code at the same time.
[00:19:53]  And suppose there is a dependency of
[00:20:00]  of one thread on the other thread when they will release then then other thread will acquire that that resource so
[00:20:08]  that dependency will will will make the race condition over over the over each other so that is a race condition in the system
[00:20:20]  Ok.
[00:20:33]  So let me just give you one more coding question.
[00:20:38]  Como é que é?
[00:21:10]  Can you solve this question just a second?
[00:21:16]  Yes.
[00:21:20]  You are given a zero index array of integer numbers of the index you are in sequence in the index.
[00:21:25]  If the index numbers I represent the maximum length of forward jump from the index I, in other words, if you are at index I, you can jump to index I plus J where
[00:21:40]  I can see the number of students to reach the index n-1.
[00:21:52]  means i have given a array of zero index where we have given each value
[00:21:59]  each value means it it is the value how much how much the person from that index can jump to the forward yes sir only forward or backward also
[00:22:14]  No, no, no, backward.
[00:22:15]  Okay, sir.
[00:22:17]  In other words, I have to just find the minimum number of the person can take to reach the n minus 1 index by using the maximum length of the given array.
[00:22:32]  Can you explain it again?
[00:22:34]  Means sir i have to find the minimum length minimum number of step jumps i need to take so that i can reach from the index to n minus one index and for the for the given condition that uh for if i will
[00:22:48]  if i will add any i at index then i can jump maximum maximum length of area of i area of yes yes yes
[00:23:00]  Sir, can I have a copy?
[00:23:05]  Ah, é isso aí.
[00:23:07]  Yes.
[00:23:13]  If I will maintain, maintain.
[00:23:19]  Maintain just a variable
[00:23:24]  and move forward.
[00:23:37]  Yes, sir.
[00:23:38]  What I am thinking is I will take a variable which will
[00:23:48]  swing the maximum step.
[00:23:51]  It can move from the current position.
[00:23:54]  And I will take another variable.
[00:23:56]  Which will correspond to the summation from each index means for each index how much I can move forward and I will just
[00:24:10]  take an answer variable and whenever the farthest position goes
[00:24:24]  goes beyond this summation till that index, then I will just increase the count.
[00:24:53]  Yes, sir.
[00:24:54]  I am thinking is I will just maintain two variables.
[00:24:59]  First, I will just add the current value to the variable and one variable will be the value which shows that how much for this I can move.
[00:25:15]  From the current index.
[00:25:18]  And whenever I will make a condition that if the current index is equal to the summation till that index, then I will just increase the
[00:25:32]  answer and I will assign the furthest value to the current position.
[00:25:38]  Like for the current position, how much I can move forward.
[00:25:44]  This I can use to solve this problem.
[00:25:47]  So how will you ensure that it is the minimum?
[00:25:52]  that whatever step whatever sir i have taken the condition that if the sum till now is equal to the current position means now i i have to move forward otherwise uh
[00:26:06]  i have no uh left jump so that that will ensure that uh that i will i will take the minimum number of jumps to move forward
[00:26:23]  Sir, can I code on someone, on some ID?
[00:26:27]  Like, the time is going to up.
[00:26:33]  Like, you can code, you can code.
[00:26:35]  Okay.
[00:26:36]  You can just pseudo code it.
[00:26:37]  I'll just go through it.
[00:26:55]  I said, if I am like,
[00:27:02]  Suppose this is the array of length n.
[00:27:10]  I will just take two variables.
[00:27:15]  First, the current value, then the other will be far value, is how much farthest I can go.
[00:27:25]  I will take a pointer or answer, sorry, answer, and I will just move.
[00:27:36]  I will just add.
[00:27:39]  On each part i will just uh add the uh the i will just check how much far i can go this far means uh far till now and i plus
[00:27:53]  and far from from the current index i will just update my farthest value from the current uh current index and if i will meta condition that if i will meta condition that i of j like
[00:28:08]  is equal to how means from the previous step
[00:28:14]  from the previous jump which i have taken till now if i have reached the point where i have to take any extra jump then i will just increase the answer and i will uh i will update my
[00:28:30]  for this value and I will just
[00:28:36]  Return the answer.
[00:28:44]  Yes, sir.
[00:28:44]  I will just...
[00:28:47]  Okay, sir.
[00:28:49]  Can you just run it for the first two test cases?
[00:28:57]  3, 1,
[00:29:10]  If I will take a
[00:29:13]  Current?
[00:29:15]  Here only?
[00:29:33]  Sir, for the 0th index, I will take current equal to 2 and for this, sorry, not current, it will be 0 still till now.
[00:29:45]  I will move for here, I will update my furthest position to 0 plus area 5 means 2, I can move furthest to 2 index or 2th index from the current index.
[00:30:00]  I will meet here that I equal to current value then I will increase the count it will increase
[00:30:11]  then and I will update it to 2 again I will move to 3 then for this will again get updated it will be
[00:30:24]  1 plus 3 equal to 4.
[00:30:28]  And I means 1.
[00:30:31]  1 is not equal to current since current is 2.
[00:30:34]  So, it will not execute.
[00:30:37]  Again, I will come for 2.
[00:30:41]  Then, far will be max of 4, 2 plus 1, 3.
[00:30:48]  It will again be 4.
[00:30:50]  And now I means 2 equal to 2.
[00:30:53]  Then again answer will increase by 1.
[00:30:57]  It will be two and our for this will be four.
[00:31:02]  For 4012343 again what will be max of far like 4 comma 3 plus 1 4 it will again be 4 and again when I come here then it will
[00:31:16]  it will not execute again since is 3 and current is 4 now again I will come for come for 4
[00:31:26]  then 4 plus 4 is equal to 8.
[00:31:28]  I will update 4 equal to 8.
[00:31:30]  to a part is equal to 8.
[00:31:34]  But sir, I have to reach.
[00:31:40]  N minus 1.
[00:31:42]  So, sir, I have to say, I will have to just take these values only.
[00:31:50]  Yes.
[00:31:51]  Yes, sir.
[00:31:52]  It will be these values only.
[00:31:56]  Okay.
[00:31:57]  Yeah, sure.
[00:31:59]  Thanks.
[00:32:00]  We will let you know.
[00:32:02]  Okay, sir.
[00:32:03]  Okay, sure.
[00:32:04]  Thank you.
[00:32:04]  Bye-bye.