Love the way how you also provided what the expectation of the interviewer at that particular time was. Really helped in understanding how an interviewer thinks. Thank you :)
Hi Keerti, thanks for putting this type of content out. You actually looked like one of the interviews who are serious as well as 'nice' and helpful' at the same time giving space to the candidate to express themselves. The way you have edited the video giving references to what needs to be done when and the approach to be followed is also very helpful ! Thank you Keerti !
Great , specially awesome from Nishant, very few people will have the courage to give mock interviews publicly, that too after being selected at Microsoft, shows that he is ready to accept his mistakes and keep learning from whatever opportunity he gets. And interviewer was really helpful 😅
This is absolute gold. I urge you to please make an entire playlist with different incoming SDE's so we can see how different people handle the process . ❤️❤️
There is no gold in this. The first question was a failed attempt to display irrelevant graph and backtracking approach before finally doing a greedy solution which he couldn't even explain , the second question was just plain wrong, the solution was to use binary search. Gold my a**.
bro same , i didn't even not get clear understanding of 1st ques and solution he does , finally sigh of relief , that it is not possible to attempt best in interview everytime
Wow . This was an amazing mock interview. I myself really have come to know about lot things by watching the whole video which would not be possible if I would just had read some interview experience. Your videos are literally good . Keep uploading this kind of content 😊
video is to the point,the best thing i like,no unwanted discussion in starting and last ,they just start a video by solving questions and end up with conclusion.👍👍 keep it continue !
His approach for the second question was wrong, he should have divided the number in 2 equal parts for even and (x-1/2, x+1/2) in case of odd. His approach will fail for A = [1, 100] k =1. he will give 99, but answer is 50
honestly, i was surprised with this approach. I initially came up with the optimised approach but then after following the video for first half, i was like, maybe this approach is better and the moment you guys came to the optimised version, i literally paused the video and gave a pat on my back :p Good content. Learnt a lot from this.
I am a second-year student. I kept trying it alongside and taking hints whenever you gave. I really enjoyed this. Please upload more such mock interviews.
Wow this was definitely one of the best mock interviews I've seen so far. 😂 I was pausing just before Nishant bhai and was trying to answer the question given by mam and lol even I did the same mistake of not properly reading the 2nd point. 😅😂 Yes mam, actually she even gave the time to read the question and also explained it a bit also told us that we didn't read the second point I don't think some Interviewers will be this patient and supportive tho
This was an amazing video, shows how things actually goes in a real DSA round. I think it will be helpful for those who are new and unaware of this. Also, will help as a guide to boost confidence before any actual DSA interview.
it was a fantastic interview given by nishant to u his concept clarity is outstanding and as an interviewer urs as well . i liked the way u r connecting with him in between the time when he was answering . superb guys .keep going keerti and nishant
Good one. this kind of interviews will give us path to focus more on problem solving skills. Thanks for sharing. I also follow Nishant channel so I also wanted to see how he performed here. 😅
He definitely didn't answer the first question properly... As it's a straight forward greedy problem and he is thinking of doing graph, backtracking and all. If this is a interview he will be definitely rejected!!!
how can u directly conclude its a greedy prob, unless u first brainstorm the solutions possible and understand the core of the problem. He was really good to do that.
@@harshavardhanranger yes you need to think for a while, but not 30 minutes to come up with the approach after so many hints... I am not opposing nishant bhai, ppl sometimes may stack with the problem, but just want to say it was not up to mark performance to clear interview if it was an actual interview...
Thanks for this video keerti. It would be grateful if u could upload this type of content frequently. I am also from NIT CALICUT and as your junior i am really proud of you
These videos are so informative and insightful. Just one thing after 2 months into leetcoding, whenever there is a term/relation to maximize the minimum or minimize the maximum, the first approach should be to solve using Binary Search! I guess it would even work in the second problem.
This video was soo good ma'am. Me as a 1st year student of CS, i have an idea of how mock interview would be and how we can talk and prepare for it. More mock interview and more questions plz would love it
An Optimized solution for the second question would be to use binary search. the range of possible answers would be [1 - max number in given array] . we can just apply the binary search on this range and for each mid which we encounter we will just check whether or not it is possible to get that mid as the max number. If it is possible to get the mid as max we will check in the left part of the array and store the mid as ans, and if it is not possible for us to get that mid as max in given array we will check in the right part. we will do it until low < high. If low >=high we will come out of the loop. I hope it does make sense 🙂🙂
@Satish Sutar I am aware of that. And I'm a third year student from a Teir 3 college and I'm trying my best to change the fact that product based companies overlook Tier 3 college students.
We can solve the first question using sliding window technique by take the window size = k and checking the ith and kth position for T and P. If ith and kth index is T and P, increment the count and move the window to right.
the approach for first program is very simple it is a greedy problem where you just need to catch the theif to your leftmost in range so that the distance between theif and police decreased and you can catch maximum thiefs.
The first question was very simple, 2 iteration (forward and backward) would have done the task. The question would have been solved for not more than 20 mins.
That seems to be a greedy approach which won’t work in second case since if P at index 2 catches T at index 1 then P at index 3 won’t be able to catch T at index 0 since distance is larger than k=2, so I don’t think it will work.
Used the concept of Stack to solve this. Like I used a police stack to store the indices of Police and another stack for theif. I used this because the condition states that each police can only catch one theif which indicates popping both the indices from each stack because they are of no use later. So the time complexity would be O(n) and Space complexity is O(n).
I think Nishant bhaiya's approach for the second question will fail for the testcases like 2,100 and k = 1. Here Max(A) = 100 , Max2(B) = 2 ; So , C = 100 - 2 = 98 ; after inserting in priority queue we get {98 , 2 , 2} in priority queue. Since we have already exhausted our k(k was 1), our max answer will be 98 , but ideally it should have been 50.(If we had simply divided it by 2).
A very simple solution is For each policemen, tabulate every element at k distance from him thus their positions are already known as index_of_police + or - k . And then for each of them , try marking the left thief first and then if left DNE, then mark the right one. Once a T is marked, don't touch him again.
Awesome video, get to know about really good perspective of how to approach solution and communicate with interviewer. Please make more videos like this!
One genuine thing was..it was not scripted. Most of the times in such playlists interviewer already know the soln and act first 10-15 min and then come to brute with 100% confidence ..This one was completely done by him at instant ! Nice work guys .
PYTHON SOLN FOR QUES 1: lis=[] n=int(input()) for i in range(n): lis.append(input()) k=int(input()) count=0 for i in range(len(lis)): if lis[i] == 'P': for j in range(len(lis)): if lis[j]=='T': if(k>=abs(i-j)): count+=1 lis[j] = 'c' break print(count)
Personally I feel that if the interviewer is explaining the question, he would sometimes hint the solution or maybe an approach to solve which can be a boon for the session. Corrections are always appreciated.
From my experience with micosoft. Don't think and find solutions yourself. Those interviewer wont accept your own solution. They want exactly what is on the internet. Spend time in look all the websites and Better memories them as they are in Internet. 90% percent you will succeed.
Max thief count problem solution has been made a little complicated by Nishant as I solved it with simple nested for loops and one Boolean array which will set an index to true whenever a thief is caught and based upon that Boolean value of thief index I can move forward. Finally I can count the total true values and that is the total count of thieves caught.
one thing that i understood from this video is that , interviewer don't want a correct solution with optimized space and time complexity in the very first try ,what he/she actually want is the discussion of 40 to 60 minutes on the question. He want to analyse how the candidate handle the situation when he got stuck or how good is he in grasping the hints . Thankyou Keerti and nishant for this video . 0_0
#Keerti Purswani thanks for letting us to have Idea about interview session and also to expand the horizon. a humble request to you to share your knowledge with us regarding learning d s algorithm and programming questions so that we can take start early Doing the great work keep it up 👍👍
This is very informative . I have been practising dsa from a long but yet not sure if i am ready for interviews or not. But this video really help to realize where i stand now. Thank you di ❤️
First Question with O(n*2) brute force approach inputarr=["T","T","T","P","P","T","T","P","P","P","P","T","T","T"] temparr= [0 for i in range(len(inputarr))] k=1 k=k+1 count=0 print(temparr) for i in range(len(inputarr)): if inputarr[i]=="P": for j in range(len(inputarr)): if (inputarr[j]=="T") and (abs(j-i)
second question approach will fail for tc [ 8 , 12 ] and k = 2 According to his approach answer will be [ 8 , 8 , 4] -> [ 4 , 4 , 8 , 4 ] so it is 8 whereas answer will be 6 [8 ,12] -> [ 8 , 6 , 6] -> [4,4,6,6]
His approach won't wok for [7,17], K = 2 as per his ans, we will do [7,7,10] then [7,7,7,3] and it ends. But we can get [7, 6,6,5] then [3,4,6,6,5] and we will have 6 as ans
in 2nd que , his approach is wrong Ex:- {7,17} k=3 According to me :- 7, 6,11 k=2 7,6,5,6 k=1 4,3,6,5,6 k=0 Max = 6 but according to nishant 7, 7,10 k=2 7,7,5,5 k=1 7,7,0,5,5 k=0 Max = 7 So he was wrong.....
A real mock interview Di...!! You really scared at some moments 😅😅... Also di he already declared vector and you were asking in which language will you be coding...😁😁 I am going to follow you greatly for my internship prep and placement prep🤟... And di one more thing if you could answer..are two projects enough...to display in our resume...of a webapp??...it would be of great help if you could let me know🤟🤟
Why am I scared though 🤭🤭 He is the one interviewing 🤭 Han I know, I wasn't expecting him to code in C++, plus I just asked so that viewers can know 😅😅 Depends on the projects and the level of your contribution 😊
My approach -: 1.take a police array and thief array and store their corresponding index. 2.iterate on police array and if absolute difference b/w each element
Can you do some frequently asked leetcode or hackerrank questions and the approach to solve them, what data structures should one use and why. Also some problems do not have a straightforward time and space complexity, so how do we tackle those. And keep going this is absolutely great and very helpful.
I liked the questions. There are lot of mock interviews floating around using the easily available/popular questions.Definately that's not the case here. Where you got these questions?
In first question, bhai kahaan chale gye the recursion dp pe. My approach was first to try simple greedy solution. And I came up in 10 mins only. Personally, I prefer if greedy is not working, then only move to trying all possible combinations.
Really good one, keep going..... You finally smiled at 32:04, till then you kind of giving a confused look :) ... I also can hear some background noise/feedback, it could be a fan sound or something.. not sure..
Cann't this question be solved using sliding window and HashMap 1.Take a window of K+1 items and maintain number of police and thiefs. 2. Find min of these values and subract with count of both police and thief. 3. Slide the window by 1, (reduce the count of the item removed from window , if any left and add new item) Keep doing this till window reaches end. Correct me if this solution won't work .
Hi Ankit, If you notice, he is doing the same thing finally. Instead of saving two vectors like he did, you are using hashmap And he used 2 iterators (2 pointers) and you are calling it window technique. If you have watched my window sliding technique video, I have explained how window technique is same as 2 pointer technique. Hope I was able to explain 😊
thank you so much for asking the interviewee to speak in Hindi, I really wanted to know the solution and not miss anything cause I am not good with Hindi.
@5:34 in that how the ans is 3🙄? It might be again 2 According to me bcz 2st index police man can catch 0th index 'T' 3rd index police can catch 1st index 'T' And last index police man would not able to catch any 'T' So if I am wrong plz explain me
I am thinking the same😂😂 the ans is 2. Edit- after thinking some time, he is correct because the policeman in last position is able to cath thief in second last position because he is less than 2 position far from police i.e 1 postion and in question it listed the a policeman cannot catch a thief who is more than k unit i.e 2 unit
the best and worst time complexity of Nishant approach is O(2*n) as each index is covered 2 times(One when creating Thief and Police array and then iterating over them) I have coded this problem with queue and its worst time complexity is 0(2*n) as each index is covered atmost 2 times.
I really wanted to know because hype of placement is so much that even if the problem comes easy it's like unbelievable for our minds like is it true.... I am solving that 450 DSA Questions by "Love Babbar" bhaiya and sometimes it's like unbelievable that these questions come under company Questions....
Exactly, for left to right we will only consider the thief on the left of the police and for the right to left we will only consider the thief to the right of the police.
