MICROSOFT Coding Interview | Getting Interviewed by a Microsoft Engineer
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- Äas pĆidĂĄn 26. 07. 2024
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The video contains following parts-
0:00-0:40 - Intro
0:41-1:35 - Question 1
1:36-4:01 - Question 1 discussion
4:02-05:17 - Crio Sponsorship
5:18-13:21 - Question 1 discussion
13:22-16:58 - Question 1 Code
17:00-17:46 - Question 2
17:47-25:25 - Question 2 discussion
25:26-35:56 - Question 2 code
35:57-36:20- Closing Notes
Hi, I am Fraz.
I am Software Engineer working at Curefit and I love teaching.
This is my Channel where I upload content related to Interviews and my personal experiences.
âš Hashtags âš
#SoftwareEngineer #FAANGM #FAANG #DTU #engineering #internship #college #DSA #Freshers
It should be row++ instead of col++
Yes thanks for the correction
Yes same doubt raised to me thank youđ
Yes i also found that error
Moving backwards increments the rows,?
â@@ankushmudgil9352 so smrtđ
really helpful more similar videos would definitely boost the confidence before campus placements
Sir first of all thanks to you and Kushal Sir to feel the Enviroment of a Real Interview and I can solve all the 2 Questions so I'm Happy that my Efforts are worth it!!
That's great đ„
we can also use a base case if(targ < a[0][0] || targ > a[n-1][n-1]) return false;
Much needed videos like this!
Just want to thank you for this!
Telling the approach in an interview is a skill, thank you for guiding us about how to give an interview!
Thanks a lot for appreciating âșïž, hope it helps
I think we can reduce the for loop iterations a bit in case we have many numbers by having a condition "nums[i]
You can treat whole matrix as 1d array
Where low = 0, high = n * m - 1
And do binary search
We can find row and col like
Mid/m , mid%m res.
Just heard your approach and tried to code myself on LC, and it was correct in the first attempt.
for the first question there is a faster approach , you have to use the sorting more efficiently, there is this O(log n+m ) solution :
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
right = len(matrix[0]) -1
bottom = len(matrix)-1
top ,left=0,0
def searchHelper(matrix,top,bottom,left,right,target,visited):
if top > bottom or left > right:
return False
mid1=(top+bottom)//2
mid2=(left+right)//2
if (mid1,mid2) in visited :
return False
visited.append((mid1,mid2))
if matrix[mid1][mid2] == target :
return True
if matrix[mid1][mid2]>target:
return searchHelper(matrix,top,bottom,left,mid2-1,target,visited) or searchHelper(matrix,top,mid1-1,left,right,target,visited)
else:
return searchHelper(matrix,top,bottom,mid2+1,right,target,visited) or searchHelper(matrix,mid1+1,bottom,left,right,target,visited)
return searchHelper(matrix,top,bottom,left,right,target,[])
using recursion is not always a good thing, but yes it can be optimized to 1 more layer using iterative approach.
For the first problem, is the solution with binary search better?
We can start from the middle in the mattrix, like from the centre of the quadrate and based on the value get rid from all left bottom values or right upper. Than define the middle of new quadrate and repeat a procedure. Time complexty in this case will be O(log(n*m)) which is much better.
Nicely explained , both problems are from this month's leetcode daily challange đđ
Thanks âșïž
hello ankur sharma bhai ek baat puchne thee pls are u there?
Great video of the mock.
Just one highlight point, clarifying questions can be better, for example in triplet question you could have asked do i have to return all the triplets or just one, or what return value is expected the triplet values or its indexes or just a boolean.
keep doing good workđđlike each & every video of yours. Amazing content, u r providing a great guidance to each one of usđ
Thanks a lot dolly
10:40 @fraz the reason you are giving that you will be able to move in two directions is a assertion not a reason. Reason is you can stand either on four corners or in between the array, inbetween you will have conflicting choices as incrementing row or col will result in bigger values and vice versa but when you start from top right or bottom left you have two non conflicting choices, if you go left you decrease, down increase and vice versa foe other situation
Exactly đ.
This is the reason why we can only start from top right or bottom left.
@@mohammadfraz The way you took it man hats off, shows why you have so many submissions on leetcode.
Max possible min time complexity is
(Logm+logn)
If we modify the binary search
i really liked this mock interview please upload more video like this it helps to prepare
bhaiya can i crack the google and microsoft beacuse recently i have started learning dsa from scracth and i am in mid of the 5th sem but i am facing problem while solving questions in optimized way most of the times i watch the solution
this playlist is best
Please continue mock interview series
inplace of col++ will be col- - And similarly inplace of row-- will be row++
1st ques is asked in my one of interview in my placement session đđ
fraz bhaiyya oo first question ka approach is really super and it's easy to understand keep sharing this knowledge bhaiyya
I recently practiced both the questions on leetcode. You did it very well. I think in the last question . The duplicates for first loop i.e for i should be the check of arr[i] and arr[i+1]. The for loop itself will handle the last increment.
Yes i was doing about to the same after discussing space and time but then we concluded it
Can you send the leetcode link for these questions ?
We need more videos like this
Thanks âșïž
Wow bhaiyaa great video....
The elements to be unique was the first thing clicked in my mindđ
Thanks âșïž
Great conversation with you Fraz, Excited for our next mock interview on my channel. â€ïžđ„
Yess đ„ , excited for your turn.
You could have given your detailed feedback at the end of the interview then only it will be helpful for the people who are preparing for interviews.
Like always ask clarifying question before start discussion the solution.
I am also taking interviews at my org and find that candidate doesn't ask clarifying questions and also they just jump on coding and doesn't discuss the solution before writing the code.
These are my expectations as an interviewer:
1. Candidate should ask clarifying question before discussing the approach to solve the problem.
2. Candidate should confirm with the interviewer if the approach looks good then only start the coding.
3. Candidate should think and code loudly so that if he/she get stuck then interviewer can give some hint.
4. Always try to write optimized code.
5. Never give up in an interview and seek hint/help if get stuck somewhere. don't seek more than 2 hints.
6. Consider edge cases and dry run the code with at least 2-3 test cases.
can you get some more medium level questions next time wich are some what harder to solve so that we can get a gist of how to handle the pressure in those situation where we are not getting to the optimal solution btw nice videođ
Interviewer questions bnata nahi hai kahi se utha ke hi lata hai except Google
Thank you bro for this type of content
its been only a week now of me learning java and i was able to ans the first question i have learnt till arrays and wasnt aware of the concept of binary search i am sure i will learn it in the future. i am writing this message because i am just happy happy hahaha
Need more videos in similar format!!
Okay bro
Keep doing more such interviews
Sure bro
If only the actual level of the questions would have been like these..
ques 1 coptimized approach was actually using binary search, worst case the given by fraz in actually N2 if element is near right corner
bhaiya for the second question in the 3rd while loop (nums[e]==nums[e+1])it should be e-- instead of e++.
In the first question, why not start at the middle element ?
The travel distance will always be shortest no matter where your target element lies on the matrix. If you start at the corner and the element lies at the end of the diagonal, you've spent 2X more time than you would have if you started at the middle.
Sir, I solved the first question in O(log (n*m)) complexity by applying Binary search in the last column and Binary Search in the row in which the target may be present.
It's great sir please upload this types of video
Thanks âșïž
In the first question, it should be col-=1 and row+=1, coz we are coming down to next row if target>current element else 1 col backwards?
Amazing video btw
I was asked the exact same 1st question for my Microsoft interview :)
When we will expect the interview to be taken by you on Kushal vijay's channel that will be to awesome
very informative
thank you. It is my dream company to work for. Really appreciate all your efforts sir.
ver informative
We can use hashmap with I and j variables sliding through all of them
If all elements in Matrix is unique then we no need to bother row and column and straightaway we can do binary search tree using a single pointer right. Am I right on this understanding?
In the first question we can apply that 2-d array binary search as well using division and mod as well ..
Won't that be actually better approach if it is sorted row wise and Column wise both
hmm it won't work without this constraint "every element on one row is greater than the elements on previous rows "
If this constraint was given , we can consider the 2d array as a 1d array and Bo binary search
this apply when you have the whole matrix is sorted .
Microsoft Engineer using GOOGLE Docs:
Document is not important or any paper its about the concept which is required
very interesting
Good video but even before starting with your brute force approach, you should have got some clarifications out of the way. Like
1. Are all elements positive?
2. Whole numbers?
3. Are all elements unique?
Approaching a solution without getting all the cases and conditions is a strict no. Since this is a mock, you should be very careful in what you teach others.
Question 1 can be solved with O(M+N) TC and constant space.
Please start with an introduction
Bhaiya in the first problem ie. Find element in sorted matrix ,row and col should be interchanged.
In first question else statement says row - - which means you are going a row above but that doesn't make any sense. On the other hand where are you checking that target is present or not in that row if column[lastElement]>target.
The first one could be done in log(m*n). nlog(m) is not optimal.
You are doing great work bhaiya Allah Bless You đŻ
Thank you đ
Easy Question puchha tumne , interview me humko tough ques q aatay hai :( But it really helped a lot to understand how interview should be.
Man that 3rd approach of first question was đ„đ„
This was good, just a suggestion , it would be good if the input and output is given in docs before explaining the question by the interviewer
Sometimes interviews don't give input or output. But we can always get it clarified
@@mohammadfraz hi Fraz . Iam new to this .. what is these questions. What kind of coding is this .. is this realated to DSA?
Yeee sahi tha !
Search in a 2d matrix, 3 Sum (leetcode question)
When was it said that the array is going to be sorted in the second question?
My one liner pythonic solution for question 2,
[i for i in list(itertools.combinations(list(set(x)), 3)) if sum(list(i)) == k]
P.S: This solution is only meant for fun.
Ohh bahi, ye question maine aaj hi kia LeetCode pe!.. logn + logm me ban jaayegi ye binary search se...
can we use a membership check right..!, Weather the element is present or not.
Plz bhaiya more video upload karo
Nice replica đđŒ
are these to boost confidence or microsoft really ask such easy questions ?
hi
you are having a great day TILL NOW ........
Good to see Virat Kohli solving problem. haha jokes aside, amazing content. subbed!
Do we have to use google docs in actual interview also?
Faraz Sir ye live kro phir bahut acha hoga btw it was great thanks a lot
Vo bhi try karenge
the 1st question is so easy i just solved it in leetcode if i were in the interview im like boom
Wow agar sachmein is level ka aata hoga microsoft ke interviews pe tab to mera mast se nikal jyega knew both the questions đ
Great đ„ all the best Sahil
@@mohammadfraz Thnx bhaiya đ internship season shuru hone wala hai college pe july se need this all the best
First is not optimal log(m*n) is optimal
3rd approach will be like,
Apply BS on 1st column, in log(#col) time we'll have the row where we can expect target element. Then in that row again apply BS to see if element is present or not, log(#col).
total TC = log(#col) + log(#row) +O(1)
If you see the question correctly this won't work
Consider the matrix like this
[1,3,7]
[2,4,6]
[8,9,10]
And let target be 7
Here if you apply bs on first column, you'll get the row where ans should be present is third not first ..
Your approch will be correct if there was this constraint " all the elements on one row is greater than elements from before rows"
thank you so much bhaiya
You're welcome
I don't think they will ask these well known questions in off campus interviews
1st question was good
I didn't knew they make you write code in microsoft word? i thought we could use normal IDEs.
Bhaiyaa i was able to solve first question đ
Wait hows the first q solution correct, he didn't even perform binary search right?
wow bhaiya bhut accha experience diya like real interview
It was real only
@@mohammadfraz yes bhaiya
col-- and row++ instead of col++ and row--
does top companies like microsoft accept candidates with extended education..as in b.tech completed in 6 yrs .
It should be row++ and else col--
you should have got some dp in the house!
Fraz bhaiya plzz bring some podcasts videos also
Sure bro
make a video who take a job from non tech branch
This is a cakewalk question. Are Microsoft coding interview questions this easy ?
Bhaiya online coding platform like gfg,codeshaf, mein code kese likhe, start, end kese kare kuch pata hai nahi chalta, vs code me code likhu too kuch error nahi ati please help me ans please
Easy questions practice karo, usse help milegi kafi
pls increase your sound your sound was less and interviewer sound was good
It's just because our audio was recorded at his side
sir only this type of quesitons are asked in interview only wow if its my interview i m clear it easly đđ
It should be col-- and row ++
Yes ur right , good observation
I think row and column variables increments and decrement has been swapped
Let me check
In elseif it should be row++ and in else col -- right?
@@dipayaanroy137 yes. Correct
Kem Cho faeraz bhai
when should i solve your sheet after completely learning dsa or while learning
it pls reply
While learning
Hii fraz aapko kitne coding question puche the product based company ke interview mei.can u tell
Its strange in all mock interview first question asked is of matrix, the same one..
Nahi nahi it's not like that
Bhaiya i am stuck in dsa i not understanding how to complete dsa in 1 month
Good try! Just keep in mind not to come to conclusion or provide a solution in the first minute or 2. take time to clarify and answer all questions and showcase with example before talking about your solution or complexity. just 2 cents!
But why you not use a DP in Second Question
ye questions to aapke liye easy honge FRAZ bhaia
Ye basic questions hi the