Live Mock DSA | GeeksforGeeks

Key Takeaways

foreign [Music] [Music] [Applause] [Music] thank you hi hi are you sure this course is just for 2499 yes and you know this is the only course with this expertise across the world no way yes way [Music] don't delay enroll now Geeks learning together [Music] hi hi are you sure you can prepare for product based companies through this course yes sitting at home no way yes way don't delay enroll now Jake's learning together yes hey uh what do you mean by a self-based course well the course material is available in the form of video lectures and there are multiple practice problems that come along as well so you can access all of it anywhere anytime and they are available to you for Lifetime no way oh yes way don't delay and enroll now Geeks learning together hey hi uh very good evening everyone welcome uh to another session of this live mock DSA interview where we conduct live DSA sessions where we interview our candidates uh very happy to see you all here and uh I can I just get a quick confirmation if my voice is clearly Audible and if my screen is visible foreign [Music] thanks everyone for the confirmation so let me introduce myself my name is suniti and I am a mentor here at Geeks for geeks and uh today we are going to uh interview our candidate whose name is divesh so let us welcome our interviewee all right hi divish foreign hello so how are you doing all right uh so uh let us just start with the introduction of yours hello everyone my name is okay uh the wish I think you were not fairly audible so can you please give your introduction once again father hello everyone my name is Divya Jivani I am currently in my final year of B Tech and I am pursuing it from Indus University Ahmedabad all right thanks so let me explain you the procedure how we are going to proceed so a Google doc link has been shared with you right so uh I will be writing down the question there and first of all you need to explain your pseudo like you need to explain your approach then you can explain the pseudo code and then I will share the link of the gfg practice portal and then you can code it there okay okay yeah so can you please share your screen if the screen is visible hello okay yeah yes your screen is visible can you open the link oh yeah okay so let me write down the first question that is given a binary tree you will need to print its level order form okay just give you the first test case and okay so if this is the binary tree all right so in this case the output that we are going to print is this okay so as you can see uh the first the root is printed as it is then first we are moving from left to right that's why we have finished 20 30 and then we are moving from right to left okay that's why we are printing 60 40. so this is why you need to proceed for the entire tree and you need to think the traversal in the spirals all right okay man so if we have the fourth level then we have to go from left to right okay exactly and uh if we have the null binary then we have to return node yeah exactly you have to return answer in a vector oh okay so here I have to return empty Vector right yes for starting we can start from root not here which is 10. then we will go level wise we can write the level wise in using BFF algorithm then we store it in Vector then if it is the level is the odd then we go from right to left and the if level number is even then we go from left to right okay you can write it down oh but and so if the level number is hot then we go from right okay uh one thing you are starting your level number from zero or one for one okay so 10 is at level one right yes okay if uh and before from let's go this is my Approach the time complexity of this or we go and and the space contact complexity is also the view taken the space for people of n okay so uh how are you traversing it I mean you know by what process by uh doing BFFs uh like uh first I will put a root node in queue okay then uh I will move further here uh if the children uh all Lords uh uh I cover which is which are in the queue and then I will tell us again the old notes I am currently standing at level two right then we have 20 and 30 in our queue right okay and then we will uh we put out 20 and the put a student uh in queue like 40. and then in second iteration uh we put out 30 and we put a 30 in the Q okay tell me one thing when you will be traversing from right to left so in that case the right note will get exerted first inside the Q right from right to left yeah like suppose when you are traversing 30 20. okay yes yeah so when you are traversing that so in that case 60 will be inserted first and 40 will be inserted first this 40 will be inserted later right no no no I will uh Traverse the in a same way but I will reverse the vector okay you will reverse a vector traversing you will do as usual and then you will reverse right yes if the level is the odd then I will reverse the vectors if it is even then I don't have to move okay so you will be creating a temporary Vector for this to like store it and then reverse it like okay so all right uh this approach seems fine to me uh you can you write down this pseudo code okay we got the root of energy right sorry I got the root of binary right for uh yes yes the rule is given to you yeah then I will start from you not expected so I will um and I start harvesting okay first I will take uh the size of the 2 is a taken because of the we are traversing level wise we don't have to like extract 40 also right that's why I am taking size that you currently here yeah so then we can get the size so of the level current level in the second level we have size of two right that's why I'm taking and I will start traversing you I think all right just a second image can you please consider keeping your laptop at some stable position because I think you know uh that is not very stable and like okay type of 50 000 more front element of a queue will be a text back out and uh will pop out that element now if current node has its religion so we have to extract the left uh children first because we are considering the left to right manner firstly then if the level number is given or the code then we will reverse it let's go upper button okay I mean right now you are doing uh right now you are doing a normal level order traversal right the the like a level number is the odd then I will um actually you don't have to return a 2d array you need to return a 1D array for the front node has the left children then we have to push it into okay okay good thing now uh our Vector X here currently we have the current level in left to right manner that about 60k for this is the binary Vision it will return through its number in the current number is port then we have to push the videos of current vector reverse yes but the face is hello oh yes I'm listening yes I think this is fine is it okay okay just possibly let me have a look for anything okay all right the approach seems fine to me what is the space complexity of this code space complexity is a like a queue is taking a big of n space and the vector X is taking bigger that's why uh we go two ends uh the spare complexity will be if you go up to 10. and speak time thank you and time complexity will be or we go of n you are reversing the vector right you are traversing the vector like inside every iteration yeah yeah so time capacity will be on the 2o into n okay all right I'll share the questioning with you okay before moving ahead uh you have used generative statement right if level number and one to check if the level number is odd or not right so can you tell me what is special about the binary representation of odd numbers so that when we do it and with one we get one as answer yes because if any number is powered in a binary representation it have the zeroth bit set like uh if we have an example in three we have uh which binary represents as one one right foreign all right uh I have shared the gfg's practice portals link with you so can you open it and code it there awesome okay function will return okay ready is there or not one edge cases all right I guess you need to insert the data inside the temp as well right foreign and also if I remember you were starting your level number one right okay thank you thank you okay uh submitted bro you should do companion run first right all right uh so the code got accepted so okay that is correct it means but all right tell me one more thing uh here you are doing this you know reversal thing inside the traversal right so uh is there any method by which you can avoid this reversal thing yes we can avoid this thing if uh we like I mean that is an additional operation that is taking place right so if you can try to avoid it I am at uh even level then I think that is not possible okay so I mean this is the only way by which we can do this question no they can do also by but it will take space extra if we do it so extra space you are already taking like extra space you are already taking by using this additional Vector right the stamped vector so here what you're doing that you are using face as well as you are doing an additional operation so maybe you can take additional space and avoid doing this additional operation yes here we can do it from a dff like a we pass the answer array and uh if the level number is even then we go for right first and uh left afterwards and if the level number is current level number is odd then we will go left first and right after DFS approach okay uh can you write down the pseudo code for it okay can I write down in this yeah okay if you are comfortable with it then why not okay in BFS I will first like declare the answer and the level number one okay now by first okay okay I'm writing the function of BFF will return nothing in which I am storing my answer right okay you you don't need to write down the full code you can just write down the pseudo code it will just save us some time level number okay or and we will uh left children yes left Children First and if the level number in E1 like if we are standing at level number two then we have to go uh write Children First right right okay so means inside the BFS thing you will be doing a DFS as well right no no no no no I am doing only DF okay you are using only DFS all right yeah in this I will do our operations and I will insert the answer all right yeah so okay the approach seems fine we can use this as well all right uh so well done so let us move to our next question and that is I'll write it down okay before moving ahead uh I was going through your resume right and I just uh saw your code forces profile lead code profile your specialist at code forces right and a guardian athlete code so why did you mention that in your introduction foreign okay so uh you were not audible I'm not sure if the issue is from my end or if it's from your end so uh audience can help us with that were you able to hear what he said can anyone in the chat section please confirm hello yeah so why did you mention it like I thought of including it but uh I would like to Let It Go and uh I am expert at code forces and not Guardian but I am knighted Athletico all right that is yet a very big achievement right see whenever you are introducing yourself in the interviews what happens like we try to you know express our qualities we try to show off ourselves right like all the good things about us so you know you must be venturing those things in the introduction as well that leaves a good impression right so just uh that's one point keep that in mind I mean that is a big achievement and you didn't add it so just keep in mind that you you should be adding it right yes okay so okay let us move to our next question that is you are given a positive number X print the largest jumping number smaller than x and for it smaller than or equal to and then what is the jumping number JN difference of adjacent digest is please is equal to one okay so for example if X is 10 then output should be 10 as well and if x is 50 then output should be 45 okay so is the question statement clear to you yes like if the value of x is the value of x is always greater than 10 or uh smaller than 10 also value of x can be anything starting from 1 to C10 is to the power 9 all right and every single digit is considered a valid jumping number okay yeah so here the value of x is nine then answer will be nine right yeah if the value of x is 17 then we have to use the answer will be 12. yes yes right yes because the difference must be one yes oh first I like I'm observing that we checked in X or X that uh if it has the uh different disease and the difference is already one if there is one then we'll be fine to return the app for less than 10. because if we have 3G three digit number like one two one then this number is also a jumping number right yes can you tell me the range of X like 1 2 10 raised to the power 9. all the integers you pay out the largest number uh like a V of the we have to return the closest number of uh project number which is jumping number to the x yeah I mean if it is largest then it will be closest to X yes that's why we can start from our index number number yes yes number itself and uh we can go till the uh till when the app will be fine the jumping number okay yeah the current number is jumping on I will check it uh like a week I create the a separate function for it and check that I convert it to the string and then checked it if it is jumping number or not because I consider two elements at a time and check the difference what will be the time complexity of this code X can be vary from 1 to 10 just to nine then uh order of x and uh the 10 raised to 9 then that is this very small number but we have to consider it that's my uh nine into uh order of okay yeah nine is for digit like if we have the number of 10 or No 9 right and we also Traverse nine digits in one go for the separate functions which are used for checking and uh we have to check for uh 10 raised to 9 numbers are you getting my points yeah so I think it'd be a quadratic time complex 9 into order of x only because we are going from X to the smallest number okay so the maximum length is going to be nine all right okay yes that's why I taken the nine so it will be 9 into order of and like yeah whatever is the integer value yes so do you think in the worst case I mean let us say if the integer value is 10 raised to the power 9 say something like that so do you think in that case your code will be optimal no no this is not optimal this is the brute first one okay this is the group first okay so can you uh can you write down the pseudocode for it or do you want to proceed directly to the portal no no no no I can write like I will create one function step number four this function will return the number is jumping or not okay in this uh I will create a number it is constant function then I will go uh and for every uh time uh I will check the function is true or not if it is true then I will return that number before just uh on Spanish Road for any number else it will do so I also like one Edge case is there because if and this is a less than 10 and it will always less than or equal to 10 right yes because this is the first one uh you know hello yes yes proceed now uh five days we can observe that we have to find the process number of f then we go from index number like index digit and we will uh minus it till the number we get so difference will be not one piece number what are you trying to do for me posted okay are you speaking something or um four five uh ninth week then uh these have to be answered all right so this is the project one which is jumping number hello okay so like uh is there any intention behind taking so large numbers or maybe we can do with taking small numbers as well okay I have got the intuition like for every position we have nine numbers so uh if we are standing at uh the uh like we are in indexing gate properly you know one two if we are standing at a index file right then uh we will try to then we will try to do the digit uh at difference one right if the current digit is 9 then the one possible digit is only eight so and we have to uh so we have to like we need a smaller number that's why we cannot get eight that's why uh we have to reducing the number nine eight eight so in first iteration we will get what are you trying to do like uh I'm traversing from right if I got this number you are then oh you are traversing your current integer like you've taken you have taken an integer and then you keep on decrementing it till you find a jumping number right and then right now you are checking that yes right now I am checking the only index digit the zero uh the fifth index digit the rightmost digit is six right okay then I will check for uh nine so the difference is three okay so I have to uh like the 6 is lesser than nine so I cannot increment it because we need smaller number than x all right so are you like telling me the optimized approach yes yes I am telling the opportunity okay so before doing that I think you left that function incomplete like where you were checking that if the current number is jumping or not the school function right you need to complete it as well but foreign [Music] okay so this is our brute force and now you can start the optimized approach so for optimize approach I will Traverse I will start traversing from right okay from this index last index I consider Io Io transfer the number in string okay okay then uh uh currently I am standing at index five and I have six right and at second uh at the fourth index I have 9. so 9 is greater than 6 then I have to like I need a smaller number then I cannot get eight right are you getting my point yes I'm getting that's why I have to decrement 9 also okay so I get eight and uh the project one is nine right uh how did the six change to nine that is 10 because the uh I got the eight and ah in eight uh I can get uh nine and uh seven for difference one right but I need a largest number right if I have decremented uh nine so it will never get uh uh never exceed X our given number that's why I have converted the into the nine so 9 is optimal because if we uh eight seven then seventh is the smaller number but we need a larger number that's why I have taken okay now uh the difference between 5 and 8 is also uh the three that is not the needs we need a difference one that's why uh we uh five and eight that's why uh we are decrement 8 to the sixth because you hello hello yes yes I'm listening uh so uh the five things because we have to decrement it we cannot increase it because we are considering optimal one uh so uh we uh we can consider four and six also but uh we have we need larger number that's why I will converted into the sixth and uh the optimal one is seven are you getting my point hello yes um your point okay now uh uh no you are breaking your voice it's breaking is now it is stable yes it is different now uh we are standing at index three at five right so we need uh the five is a and four difference is one then we proceed further now we are standing at four right we need the closest one to the zero so we get the digit 1. right hello yes the wish proceed I am listening and uh the closest one to the one is two then three then four and uh now the uh this number is jumping number because the all the difference are differences are one now here uh the one case we need to consider is like so I think the older cases are recovered okay so we can so let's see the time and yes I am starting from okay so what will be the time complexity time complexity is uh like a constant like a big of nine like we have to Traverse the now at Max 9 digits because the max number is 10 raised to 9 and uh for for this uh like we have two Traverse nine times it makes nine times because we have to consider the quarter nine Okay so yeah so I mean you have converted your integer to a string and then you are doing the operations right yes okay uh can you do this without converting it into string I mean the approach that you are using that is more of a CP approach right I mean know where it is wrong it is perfectly correct but uh can you try doing without converting it to a string because you know sometimes it's like whatever the data is given to you sometimes you know it is required that you are not changing the type of the data you should use the data as it is given to you right I mean without any modifications so if you can do it in any other way yes we can proceed greedily also like if we are standing at position one right no no we are not converting it into string right so I think that will be how you will position it okay okay uh let's see let's try to find pattern because uh if the number is jumping number okay this is The Dumping number like so take your time and if you need any hints then do let me know I'll help you with the hint foreign foreign yeah so tell me if one is the current number that you have then what will be the next valid jumping number uh two or zero oh okay okay so one is the okay two two yeah next okay it will be two now suppose you have one two with you okay what will be the next valid jumping number um yes yeah so after 12 what are you going to get to say when it was what we had 12 okay now we have 12 what is the next number that you are going to get 21. I mean uh not like that actually you know by adding a digit which extra digit can you add to 12 so that it will be still a valid jump three we can add a three also and uh four okay no no no not like that I mean uh see if it is one two three uh sorry equals 12 right so one two three is a valid jumping number after adding an additional digit and other can be one two one right yes okay now let us consider one two three for one two three the next valid numbers can be one two three four and one two three two okay okay yeah what did you understood uh we start from Gone and uh we find the uh find the all valid numbers all jumping numbers till 10 days to nine and then we also just binary not till 10 days to the power 9 maybe we just need to find till the last number I mean if x is given to us when we need to find till X only right we don't need a number greater than x yes uh three two one is also valid right warning also very yeah it is valid but we are not generating it from 12 right so what I'm telling is that from every number you can generate many numbers which are jumping numbers right so you need to consider all the jumping numbers that you can generate and you know you need to consider it till that point till which it is less than x okay okay and uh I have to return the maximum okay yeah and also you need to you will need to use a data structure here right like we are doing a certain kind of approach here if you are able to visualize it yes exactly so like this is a sort of a BFS approach right like considering all the numbers and then generating numbers from the present number right every number from every number we try to generate the uh all the numbers jumping numbers okay so you can uh you know push it back inside the queue and then you can I feel like yeah like let us say in the first level in the first level of the queue let us say we are keeping let us okay say in the first level of a cube we are keeping only first digit numbers only one digit numbers okay yeah so what you can yeah what you can do is in the next level for one you can add one zero and one two right for two you can write one and uh two three right for three you can write P two and three four right so I mean so on you can just keep on pushing back and so our new queue will look like n 12 21 23 32 34 are you getting my point yes yes so uh now this this was let us say this is the first level okay so the first step will look like this and now this is the second given of the second level will now look like this right now how is the third level will look for the third level so 10 you will push the data that is one zero one and right and for every integer you can check that yeah whether it is the maximum integer that we are getting or not okay I get it so did you get the approach yes yes all right so if you got the approach so you know you already did VFS in the previous question so I guess you'll be able to code it and as we are running short of time right now so uh okay I'm just uh pausing it here so all right um that was all about the next question and let us come to the feedback part [Music] okay so uh on the basis of this session first of all let me ask you how much would you rate yourself I think the 7.5 hello 7.5 okay that's pretty correct marking I would say so you know with respect to first question first question was the level or a privacy spiral form so you did it correct at the first go there were no any blocks here right uh you completed it and you explained other approaches as well but uh there was like an additional function that you were using so I asked you to tell me an approach that we can do without using that additional function so you uh explained that EFS approach and I'm pretty sure you'll be able to put that as well so yeah on that question you were very much clear the logic was clear to you all right all the approaches were clear to you so 10 on 10 for that question okay now uh coming to this next question you started why okay The Brute first approach was correct no doubt brute force was correct so the first thing that I would say you is that don't try to modify the data type that is given to you all right I mean yeah that is not actually preferable in from CP point of view that is completely correct okay but from this interview point of view uh you should uh you know try to not do that all right so that is there and uh so when I explain you the approach that how we can use this BFS thing so you were able to grasp it so that is a good thing you were able to understand my thoughts all right so that's a positive side and uh as you used hint so uh I will be not giving you 10 for this so okay maybe uh yeah just eight eight or seven point five for this and also one thing is that C2 things actually the first thing is that in your introduction you should have added your achievement on the lead code forces and Geeks for geeks all right I mean that are really achievements you know when I went to your resume it was very impressive and that's why I kind of selected a little bit harder questions right because your profile was very impressive so the level of Elections needed to be matched that thing so you need to add those things in the interview right that's a great thing that you have done I mean that is not something that you need to hide okay so in interview that is what we do we try to showcase our skills all right so keep that in mind and secondly I would say that maybe you can work a little bit on your communication skills because you know there will be longer silences all right so I think that that point can be improved and there is always a scope of improvement we all are learning right so yeah yeah you can always exactly so just try to little bit improve that all right your communication skills and try to speak more like communicate more and try to think out loud okay so because if interviewer is able to understand what you are thinking what you are doing then that will be easy for the interviewer to analyze your skills analyze your logical thinking okay yeah on and also like on the logic part on the problem solving part you are really great you did all that on a very decent level just for this communication skills maybe and also you know uh maybe you got shy for explaining your achievements in the interview like when I ask you to introduce yourself or whatever it was there just try to you know try to communicate okay okay yeah thank you yeah so anything from your side oh it was great session uh thank you for uh giving me this opportunity and I've learned a lot of things from this session and thanks uh we are thankfully we need to have you in the session it was a decent session and I hope you know our audiences will get a lot to learn from this session all right and uh we hope that this session will help you in your preparation Journey as well and we wish you all the best for your future thank you thank you okay bye devious okay so uh there it was the interview session with the wish and certain things that he needed to work upon was actually the communication skills right as we all saw that you know there were certain silences that was not looking pretty good so that is something that we all can try to improve to have a very good communication skills and other than that his problem solving skills were great right so that was some of the things about this session I hope all of you liked it and if there is any suggestions any feedbacks please feel free to add it in the chat section and also if any of you are interested to be a part of these sessions then there is a Google form link that is added in the description of this video do fill out that form right and then we will be having you in our next sessions there is no any such criteria that you know you should be of this level or that thing it's just that you should be good in problem solving right so make sure you are filling out that form and that was all about today's session bye everyone keep up the hard work