Live Mock DSA | GeeksforGeeks

Key Takeaways

[Music] thank you [Music] [Music] foreign 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 yes [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 hi and welcome to cakes for cakes so we are back again with yet another rung of DSA more so before we start let me introduce myself I'm sadata currently I'm an SD intern at Geeks for cakes I'm a candidate Master at code forces now without any further Ado let us just add people to the Stream hi people how are you hello sir I am good how are you yeah I'm fine too so before we start can we have a bit introduction about you yes sir I am people finally your computer science engineering a student of an attitude and I have spent three years here and I have learned technology like web development I have on them and then my technical skills are like uh uh data structures and algorithm and computer networking and operating systems and the skills and then my main interest are like solving problem solving and participating in variance contests like code forces and gfg date code and free time I would love to play chess okay fair enough so before we start let can you tell us that what are the text tags that you have explored during the learning of your web development uh node.js Express and then react HTML CSS and JavaScript okay fair enough okay so before we start let me just uh tell you as well as the audience the rules so what I would be doing is I would be sharing you a doc in the private chat okay and this would be a mock interview for 45 to 60 minutes and then I would send you the question and I would read it and as we move forward you need to start so you just tell me the approach and then as we move forward we you need to hard code it if I'm if I'm good to go with the solution fair enough so what you can do is you can go to the private chat you can open the link that I've shared with you make sure you turn off your notification so that your privacy is not Disturbed yeah I can read your name and yes it is live okay yeah many people say HTML CSS JavaScript this is the ordering okay I like this card all the same shows okay okay many people are wishing you all the best people thank you so when you are done just share your screen okay okay we have some blessings for you okay can I add it to the Stream yes okay uh actually I have shared the screen okay the screen is visible okay yeah yeah the screen is absolutely visible okay so now what you can do is you can just make it full screen if you want okay not need it so just remove this uh dark reader just disable this dark leader uh yes so the very first question is I will just type type it around okay just let me yeah okay so the very first question is a simple question so you are given an array and an element element X okay you would be given with that element okay find the number of elements less than less than x and the special thing to note here is that you are given a sorted array okay so I will just bold it so that you don't get confused so you are given a sorted array and an element X and you need to find the number of elements less than x so let so are you clear with the problem statement then we would move forward to the test cases yes I'm clear with the problem okay fair fair enough now let us move forward so the very first test case one would be given given an array like one two three four five six seven eight okay and then suppose the value of x is equals to 6 then your answer would be equal to the value how many numbers are less than six one two three four five so the value of this would be 5 itself okay yeah do you need one more test case or you are good to go with this one yes I am good to go I guess okay so okay so Brute Force approach would be like I will Traverse the whole array from starting I is equal to zero and I will check this current element is lesser than equals to this X or node so whenever I encounter first element that is greater than x so I will break them that index is up to that index is my answer up to that industry all elements are lesser than that so I will return simply uh that indices up to that element uh element are lesser than that so for example here I will iterate through this one so only lesser than six it's fine so again two is lesser than six to find so three is lesser than 4 is also lesser than 5 is also relation but it's six six is equals to the six so here I will break from here I will return this indexes because basically now 0 is it's zero base induction so zero one two three four five So my answer will be 5 over here so what would be the time complexity of the solution uh time completely will be like big off and because I have two Traverse in most case for hold array so time condition will be open okay fair enough so do we have a better solution than this can you optimize over the speed is time complexity because space complexity is already constant yes so another way like I can use binary search over here uh I will use binary search over this four indexes we have zero and then basically we need to find particular indexes that is uh function because that is greater than equals to uh X so I will check first I will check for Mid uh initially low is my zero and high uh last last index I will check from it particular that uh mid array of mid is greater than x or node if that is greater than x so I will go in the left part or if it's lesser than that I will go into the right part so whenever I encounter the first element that is greater than this there I will break okay so can you code this so let's suppose you would have like this okay so you need to return the value and the name would be fun okay and you are given an array okay and you are given the size of the array as a n and then you are given with the element of that okay fair enough so yeah we'll even now coded so initially I am working as hint answer as opposed to my n okay all element time is dreaming rather lesser than uh lesser than equals to X so after that in close to zero and uh R is equals to n minus 1. and while call call sorry I will attribute through while look and then I will calculate the mid is going to left plus right by two and I will check if mid is array of mid um it's greater than uh X that means that particular mid index can be my answer so here I will update my answer as well answer is equal to Mid and uh and now in this case right is equal to Mid minus 1. and here else right is equal to Plus after that I will return it is greater than equal to yeah so for this test cases uh but should I verify this code for this given test case yes absolutely so initially my left is at zero and right at this index 0 1 or 7th index now 7 plus 0 that is by 2 that is 3 0 1 2 3 that I will check array of four uh array of 4 is noted the next one I will go up in the else part so left is equal to plus one so now I am at here uh that left is equals to 4 and right 2 is equal to 7 so now 4 plus 7 that is 4 plus 7 by 2 that is mid so 4 plus 7 by 2 that is five so now uh mid is equals to five so I will check for five so zero uh this uh index is five so I am checking this so now array of mid is greater than x so that is at 6 is greater than five so as of now my answer is index six sorry uh index five now right is equals to Mid minus seven now again I am going for here uh now uh five to six now we have so now again with equals to five so I will check for this index so foreign so now in this case uh my right will be there and then uh so reluctable with this index four index and right will be this three index so this can this case will not go while the left is less than so I will break from while loop so I will return my answer okay fair enough so what do you think are you done with the debugging thing are you done with the testing part uh I think [Music] yes I think I'm done okay uh in case let's say uh if we have x equals to 0 so here in case we have uh uh in case if x is equals to zero so how many elements we will check for will go up to the end and we didn't encounter events in that case okay here uh here I need to write 0. initially I am assuming that I don't know any element release greater than equal to so yeah now it's fine again okay okay you are done with yours yes sir okay fair enough now take an input the input is that let us take this test case test case 2 and the input is 5 5 5 5 5 5 5 5 5 okay let's take this 3 5 okay fair enough and then the value of x is 5 itself so answer to this would be zero yeah okay so what would be the answer in your case can you please do the trial okay so okay the test case we are running is this so I would be marking actually I got my mistake here there should not be equal to 10. so that is the one mistake after that if I run this code first my mid value is like zero and then right is 2 so 0 plus 2 by 2 that is five so now our limit is greater than equals to X so it's not greater than equal so I will go in that but I'll 1. again I will check for zero so in this case also we can see that array of 0 is not greater than x so now in this case I will go in the uh in this part left is equals to me to plus one so now now my uh left and right are both uh uh one now both at index one so now again I will go array of one plus one so again I will go else part so now my left will be equal to so in this case this will valid so now my answer will not change the answer will be answer so return answer look fair enough so what is the time complexity of the solution uh time complexity would be like big of a log n because of I'm using bandwizard limit so we go plugin and the space is also constant so can you give me a idea that why does binary search has the time complexity login oh yes because of every time I am ignoring half part of the array so let's say we have a let's say we have a whole array so every time either I am checking for this meets either I will ignore this half part or either I will ignore this I have part so again let's say I'm ignoring this upper so again I'm at it again so basically at the every time I am dividing it into the half part so that's why to the power basically uh login times we will take it okay fair enough so this was a good stroke let's move to the very next problem for the very next problem I would be starting with a fresh new page okay and the question is that you are given an array okay so can you tell me what do you mean by permutation like what is the meaning of permutation permutation means like uh basically all the letters or characters basically are the numbers all are same basically they are in different order look fair enough so what we do is like suppose we are given a series of number okay there can be many permutation how many number of permutation the number of can you tell me how many suppose uh array contains n elements so how many number of permutations would be there certain numbers all numbers are unique or they might contain duplicate value also all numbers are unique so n factorial uh permutation will go okay n factorial permutations fair enough so now you are given a number okay you are given an array and what you need to do is you are given a particular permutation of that array and you need to reduce the next permutation of the given array okay so let me just tell you uh test case so suppose you are given with three to one so if you see that 3 2 1 is the last permutation just let me okay just three two one is the last permutation of the given array one two three so the next permutation of three to one would be nothing but one two and three okay do you want one more test case to be told yes okay fair enough the very next test case is that we are given an array like this okay so let me just write PC2 we are given array like this um perfect and then the next permutation of this array would be nothing but this okay a newer provider and this would be the same functional programming where you are provided with the yes okay okay so okay I'm thinking about the blue what Brute Force approach can be of this problem okay take your time it's absolutely fine so uh one thing you like will Traverse from the right side and then basically we want uh array that is just greater than this basically if it is a number one two three six five four so we want that number that just greater than this one that if it is not available [Music] foreign three six five okay yeah okay so basically we want just um greater than the number so [Music] like [Music] for one more thing I want to ask what is the size of this array what could be how are the size of that let's take the let's size of the array is not variable at the moment let's suppose it depends on the solution you have okay I don't want you to give me the most optimal one or you need to pass the test cases like you do in competitive programming I don't want that see so yes so first approach is like a very good resource approach is like I will just convert all the number into the numbers like one two three four uh six and then I will just permute I will uh put into the vector all to to uh sorry all and factorial number and then I will find the next greater l number okay it's fine but tell me why do but why do you really need to copy it to a string in the array itself yes yes yes yes yes so the thing is like first I need to just next greater so I will iterate from the uh and side and then whenever I encounter the value lesser than this one for example here 4 is greater than 5 we cannot do anything six is also greater than five so we can here we have a three three is lesser than this six so I need to just basically uh Swap this value so that the our value all value are greater than this one so now here 3 will swipe these three with this four so now one two four six five four will give them my next greater element is okay so I will just my idea I just iterate from the right hand side that whenever I encounter array of this a particular index is lesser than the RFI plus one so I will just break from here and then I will just reverse all these well you have sort this value and then rest all are same so I don't need to change anything about one comma two but I only just Swap this value 3 and then fourth and then I will sort this one all I will reverse this one this both are are fine because it always this is an increasing you would so you would sort this or you would reverse this both are completely different yes I will reverse this because of sorting login times so that's why and Logan so I will reverse into open okay look fair enough so now what you need to do is I'm going to send you a link okay in the private chat or in the dock itself and go to this link from start okay okay yeah let me just show you so does the permutation needs to be lexiographically greater than the input yes for most of the time but not all the time because it might be the case that the given permutation is the last one or the greatest permutation of that would be cyclic so I'm not replying you yeah you can just continue I'm just not asking you okay I'm just replying to the folks that are here okay so the last permutation can be like suppose what one two three the given permutation is three to one so the next permutation would be one two three okay so the ekit can be the cyclic so it does not doesn't mean that the next permutation would always be lecture graphically greater okay I think reversing the server okay okay fair enough let us see okay he started iterating from the very back yes sir uh now is greater than zero and then I minus minus and whenever I encounter if uh array of I which is lesser than array of okay sorry to interrupt you in the middle just hold command command plus okay or the control positive to increase the font size yeah no no just make it a little bit small just make it a little bit small yes yes this is fine this is fine both would fit here now you can start this 872 so it is I'm attracting from the back and then whenever I encounter of other if array file is lesser than a lesson at a five plugin I will break from it I will break it and then and then simply okay so and then okay after that I need to just reverse after the first I need to swap with the uh with the array of ith value with the very last array of n minus one and then I need to reverse the rest element uh divorce uh strong index I plus 1 this is between so we have array so for X plus I plus one two array to array of plus n minus okay and then yes and then we need to just return in this code yes absolutely so here take a smaller input take a smaller input of just three to one okay yeah okay so I think I have made one mistake over there so here we have one two three uh we don't have any array of files not lesser than audio five plus one so it continues here also it will so it will break so so we don't have any numbers how can I swap so first thing is like a all the time my eye value is zero initially now I will swap this one and then three so now the array will be like one and then two and then three after swapping this so and then I am reversing up to that element from uh or a plus n minus one so it will return okay so I think I have made some mistake probably foreign [Music] so first solve it for one two three okay we talk in general we can't just solve it for specific ones uh wool has an average equals to false and here simply uh whenever I encounter this year before breaking I will check that the has the last thing I am writing here has lost is equal to room and Air Force so I will take over here if hello I will simply reverse over here reverse array and then else I will just Swap this element of three with four so three will be there and then I will reverse here okay just comment the line 31 and try to compile try to submit it okay you can hit the compile and run if you want yes let me check on okay let me just remove me and you yeah so was the problem good yes no I'm not asking you I would just take your name whenever I ask you it's fine okay so to the audience where the problem is good what do you think okay what about others um no not really I would explain how and why just wait a sec just after the five minutes this one would become very interesting trust me just wait for five to ten minutes this is not what you think is whether this person is good with STL it seems a competitive program is always good with STL now thank you I thought honestly it is a array of integer that's why I wrote it because I I showed you another yeah I completely understand it's absolutely fun and I even told you that this would be a pass by reference not passed by value just compile and run again it is correct but sometimes it happens yeah see the stem this compilation error shouldn't happen foreign let's move forward and submit it foreign I think we don't really need to do that okay so what do you think how people performed before I start uh with this uh feedback thing how what would your rate be pulled in this interview on a scale of 1 to 10. I guess that this test case okay please just see and then tell okay okay so okay okay you are making a very small syntactical error leave it your code is fine okay that's fine now let us talk some bit so let us see your code let us review the code can you remove the output window and make it full screen the code editor part make your coordinator part full screen no there is a tab yeah yes the way yeah increase the font size yes okay so you told that has lost has been initialized with true okay yes there are some small small things that in that could be done to improve the code quality of your code okay what are the small things that you can do the very first thing is you are initializing hash has lost to true yeah okay but that would prove when last is not present yes so the naming convention is not complying with the code okay and the naming convention is absolutely very important so whenever from the next time just don't try to hurry okay don't try to hurry take some time and then think and then initialize the Boolean a lot okay now the next thing the Boolean was really not needed why because you could have initialized I is equal to minus one yes and it could never be negative and you could have checked after the if for statement that if I is still equal to minus 1 then you can just reverse it and then you can return it do I make sense yes yes sure okay fair enough now I would be removing the screen and then we would be talking about okay so after the stream please make sure that you just solve that issue okay fair enough so now let us start from the bottom okay because the bottom one question is very fresh in your mind so so time for feedback now so what I think is when you started off with the next permutation thing okay you told that you are giving a Brute Force approach and The Brute Force approach to this would be to get all the permutation find the location of that particular permutation if that location is the last one you need to report the first one if the location is not the last one if it is lying between 0 to n minus 1 then you need to report the I get plus one do I make sense yes yes so the time complexity of this one would be uncactorial okay yes yes suppose you could have initiated that I would have directly used the function known as next permutation that is given already given in STL then when could I when would I have told you that the steel is not allowed can you code it then you could have told me that Siddharth what we can do is we can have a Brute Force approach to this we can have all the permutations and then we can return accordingly okay okay okay fair enough now you could have moved forward with this permutation thing and the middle approach generating all the permutation is a very good level problem in recursion do you know that yes yes it requires a good level of concept so I was expecting that but you directly went to the most optimized one now I'm not saying that you should always go from Brute Force to optimized if you already know the optimized I'm not seeing you but what I have in mind is it just does it so what I suggest to you is that suppose you know the problem so there would be two instances that the interviewer would have the first instance is that he might think that you have already seen up this problem and you have learned that problem by heart and you don't even know the basics of it like your Basics is not right and you have just mugged up the problem okay yeah fair enough so you need to work on this that you still remember the basics okay you are a competitive programmer so a good Brute Force approach and then walking around to the optimized approach is very much well expected from you okay fair enough so this is what I can expect the very next thing was that whenever you are doing the driver the very first let's start last problem your luck was in favor and the output the input value you were trying was the input that your code would fail okay so if you go to the dock that was given to you can see that I have already copied that doc or wrote this test case in the talk I was about to go down a little bit and see this edit was made 12 minutes ago by Siddharth azra so but your luck was in favor and the Very test case for which you were doing the trial you got your but this won't happen this won't happen okay so please keep this thing in mind and the very next one is that what you can do is whenever you are explaining your approach or doing the trial so what the what do you mean by driving suppose our Mentor so I can tell you by my perspective or you're a student so whenever you are seeing a code like your competitive programmer suppose you want to start off with this binary search thing so if someone is doing the driver what do you expect him to do okay so you expect him to change the variable names also like what index it is currently at instead of just giving it overview so instead of just telling that now I is equals to 5 if Y is equals to 10 I is at index 10. it might be a scene might be a stage that you are speaking continuously for five five minutes so the interviewer might lose track so either if the code is easy you can ask him sir are you with me okay okay or you can just mention and simply draw a small variable names and change the valued accordingly okay okay do you get my point yes sir sure the very next thing was that the very next let's talk about the very first problem the very first problem you directly took the for Loop then you moved forward with the thing known as binary search and that was absolutely a great transition and it was a very smooth transition to be honest like the amount of confidence you had in speaking that yes this can be solved using binary search and the amount of control you have on the binary search code to play with it it's absolutely very commendable to be honest okay but what I expect you to do is you already have a upper bound we have so you can directly tell that yes I can move forward with upper bound then I would be telling no upper bound is restricted here then you can move forward because your primary language is C plus plus take advantage of that yes yes you'll get my point yes and you obviously are good with a steel only so it's not that you don't know just you have skipped those parts yes here okay the very next thing is that my suggestion to you would be don't get nervous okay the other person will understand you most of the time like he would try to calm you down and don't try to rush things okay it's not that the com this is not competitive programming to be honest the time frame is not too obvious so if you take time and if you provide a good quality okay and if you provide a good quality with many testings to be done and the code is absolutely Flawless then trust me it's absolutely the good approach okay so so let's talk about the overall let's talk about the overall interview okay your introduction was absolutely great and then the amount of confidence you had was commendable and then you went moving forward and it seemed that yes you have done a good amount of hard work over this coding thing okay if you hadn't been nervous those two small test cases wouldn't have been mixed to be honest that binary search error that you made that less than equal to thing you wouldn't have made if I if you were not there in not there in your interview suppose I am your friend and I'm asking you you don't have made that mistake I think so what do you think that's why okay fair enough so let so overall so overall dating to overall rating is seven from me okay but the amount of confidence you have and the code quality absolutely Flawless now one more thing which I want to add I forgot to tell you that when I started leading the binary search code you initialized with lnr so you need to differentiate the barrier that this is not competitive programming it isn't the scenario that you if you take out the code and Easy in five minutes it would be great so try to mention it either left or right or low or high instead of just L and r okay I know we completed programmers have a habit that we initialize everything with lnr so to be honest I still when I code in Geeks for geeks for lead code or any other platform if it is a functional programming and the given array is ARR and make it to a I to do it we are not comfortable I am not comfortable with ARR I to do it but when it is an interview you need to care take care of these all things sense yes sir okay fair enough so now I would quickly read out the questions for you there are some things for you okay so you can preview the screen yeah so there is Easter West okay I think that he is a friend of yours okay fair enough it's uh okay many people are very enjoyed this okay okay many people are congratulating you many people are appreciating your efforts as well as the artwork that you have given it and to be honest you deserve it because the amount of confidence you have the amount of speed you have the amount of the amount of code quality you have you deserve a pat in the back so that was pretty much from my side according in respect of feedback do you have anything to tell or ask um how I explain the things basically that is my main problem uh should I explain better or actually I need to uh more but at least you are able to understand or not okay so see the thing was I was able to understand okay but uh what you can do is you can go a bit slow okay and after speaking for 40 minutes keep a timer okay like what you can do is just after you have written the code so at what year you are in like when would you have videos okay fair enough so what you can do is nowadays your interviews would be right in the corner so now whenever you so whenever you are solving problems after you have solved the problem go go to the mirror okay go to the mirror sit in front of me and try to explain yourself only try to analyze how you are explaining code try it or just or the most easy thing would be install an OBS record it and then see it yourself you would find the flaws in your cell phone absolutely not an issue just try to think that if I didn't knew this question will I be able to explain myself do I make sense yes yes okay so what the small tips from my side would be that you little bit practice on your end and you go a little bit slow okay because you are speaking very fast without any pause or any commas so what might happen is that the in interviewer might lose a track okay and then the context of your solution would make no sense after that point of time okay then he needs to guess it from the code that you have written it okay fair so that was the thing any more further queries you have oh we are good to go no thank you okay fair enough so we would be waiting for the next 30 seconds so let us know if you so how much were you able to learn from this mock interview on a scale of one to ten quickly rate on a scale of 1 to 10. how much did you learn people what about you I have learned a lot a lot of a lot of improvement basically when I give a real interview they want to give any feedbacks but here you give that's a good for me well look to be honest what you can do is I to start it off so after you are done with the interview now you can ask him that how can I improve suppose you are giving an interview and yeah and you think that you have destroyed your interview we have many interviews like that and we know before the results are out we know that we have destroyed this interview so you can ask that how can I be better in this and I'm not I'm not telling you that there would be a guarantee that most of the people would explain but a majority of the people would be there to tell you okay okay fair enough uh we are good to go then okay so that was it give you another opportunity for an entry okay that's absolutely a pleasure okay okay fair enough so now I would be removing you so that was it from Bay pool and he performed very well in this interview and there might have been some instances that a small mistake might have been there okay in the place of in the place either from me or from him so considering we both are humans I think we are good to go that is it for today thank you everyone have a nice day