So, should we do a hard Google coding interview next? 👀 Be sure to check out the other video that we filmed on Ben’s channel, where he gave me an intermediate React interview! th-cam.com/video/6s0OVdoo4Q4/w-d-xo.html
Watched this the day after you released Clement. During an interview with a fintech startup a week later, they gave me this exact problem. I was able to remember and successfully work through roughly 80% of it. Got offered the job and ultimately rejected their offer because one interviewer who would have been my daily collaborator, was rude and demeaning. All in all, this is great information and a great video you've produced here!
Nice! Fintech companies IME can be horrible places to work because many are fundamentally banks/insurance companies at heart with tech departments. You'll be forced to appease some pretty unpleasant people in Fintech and try as they might Fintech just can't get the whole "tech" vibe good companies have.
@@Rob-vg6lw I come from a construction and ranching background; worked with plenty of A-holes. Working with A-holes doesn't make you a man. Ethics, love, and principles do. Thankfully I have those, and am now in a much richer and better environment and am in control of my future. Word of advice - whoever told you it's manly to take a miserable job over better options, was incredibly wrong. Sometimes you may have no other option, but when you can afford to take a different job with a more supportive group, do it.
I think this shows that you need to practice specifically for these coding interviews. It doesn't matter if you are a programmer or you can code or if you are smart and you have experience. Ben is all these things. You need to practice problems like these to get a job at a company that does these interviews. It's like preparing for an exam. Even if you do ok like Ben did someone will do better because he spent hours and hours practicing. So many people want to get in these companies that it is very competitive. It's just how it is
^^^THIS. I am nowhere as knowledgable as Ben in technologies that are actually useful on the job, but because I have done many similar problems, it only took me 10 min to code up the solution. I didn't even have to think, my hands just spit out code from muscle memory lol
I'm 17 and I almost always have a solution to leetclde medium or easy level problems in less than a minute. Hard usually needs some testing so that the solution is fast enough.
Almost two months ago, I started this video but didn't finish it because I wanted to solve this problem for myself. I didn't find a way to do it on my own, and both the problem and the video sat in the back of my mind for weeks. Today, after quite a lot of reading and practice (on other problems!), I returned to the problem with a fresh point of view and solved it in an hour (plus a few minutes), in one sitting, without any outside reference. It's been tough measuring my progress over the last few months, but thanks for giving me a point of reference!
You'd think that all software developers are building search engines. Lmfao. These coding questions can be solved, but the majority of problems in the software industry revolve around devops and system maintenance. Sometimes building new extensions / microservices / libraries, but these kind of problems only present themselves in my own pet (hobby) projects. Basically you can solve all of these abstract problems and still be a noob by lack of experience. Imo, time is much better spent by studying apis.
@@frankzander6234 For hobby projects yes. Majority of jobs as developer in the corporate world to earn a living requires quite a different skillset. That is in my experience.
@@kimgysen10 I don't think he's trying to progress professionally when he wrote this comment, the progress he's talking about is the progress he's made in how he thinks and his ability to problem solve. The relevancy of the project is none.
Here's my approach 1. Loop all edges - Only edges 2. If it's 0, then move to next 3. If it's 1, then update to 0 and find adjacent 1 and update to 0 until it can't find adjacent 1 At the end of this loop, you end up with a new matrix where 1s connected to edge is all 0s then finally, you subtract the new matrix from the original. Since all edges and connected cells are turn to 0, you're not changing anything, but the island cells will be 0
Seeing this level of skill, almost feel as if I am not worth my salt to be in this industry. As someone who does app development for a living, this was overwhelming and quite literally humbling to watch.
seriously? his thinking was really messy。i had watched some college kid wrote a much more complicated board game in one sit including all the gui code in c without internet access，and almost no bugs. This guy was like without thinking and just try
That was not impressive at all I don't think. It felt to me like early monday morning level of coding, or really nervous level of coding. Maybe if it was me in his place I would do stupid stuff too, but either way, he didn't impress me :)
I'm an experienced software developer but I would have definitively not passed that test. These Google interviews are all about algorithms which is something I learned twenty years ago in my studies but in my professional life I hardly ever came across algorithms again. I think this kind of interviews are pretty artificial and to pass them you have to train for it. I'm pretty sure it won't have much to do with your real work you will do.
for the "is_border" and "is_border_matrix" functions, since returning True and False over an If statement is redundant, I would have simply done a one-liner: "return ". That would tell me he's been through a number of code reviews or is part of a larger team. I've done my share of interviewing candidates as a Senior Eng at a huge retail company and even for me, a Google interview is anxiety-inducing.
Mark all the ones you shouldn’t touch with a different value. You can use DFS to do this. Go over the entire grid at the end, changing the marked cells back to one, the 1s to zero, and the 0s skip.
I'm basically new to code. I am currently learning the very basics of C++. This video is way harder to understand for me but i watched either way. I will rewatch it again in a few years and hopefully my skills are better and i can be able to understand it. Great content!!
Adding on to Ryan’s comment. How is your understanding of the video now? I’m about 75% of the way through Codecademy’s Full Stack development course and I can understand the majority of what’s going on, but still am lost in some places.
thank you for making these videos! I solved this problem by modifying the grid to have perimeter of 1s (so both len(grid) and len(grid[0]) increase by 2) and running BFS on the grid. After finding an island, all other islands are changed to 0s. Pausing the video after you explain the question and talking out loud "back" to you has been so so helpful.
Simple and straightforward - DFS from each position on the border if its marked as 1 and mark everything in the dfs as 2. Then remove all 1 from the matrix and convert all 2 back into 1. O(n) time O(1) space.
Just wanted to say a nice very fast solution I thought up to this problem: 1) Make a second matrix (we'll call it m2) of the same size filled with zeroes 2) Iterate over the edges of the original matrix. 3) Each time you encounter a 1, do a depth first search. At each visited position, add a 1 to the corresponding position in m2 and then set the value to 0 in the original matrix. 4) Once you've iterated all edges, m2 will contain all islands that connected to the edges, though none that didn't. Return m2.
@@Lorkwondo1234 just helpful to avoid having to keep track of which positions you've visited. Setting to 0 means future function calls that check a cell which has already been visited just return.
Oof i don't know what is dfs yet, will learn about it and come back 🤠 I was thinking the same approach but didn't knew how to determine if it's on the edge or not. Maybe after knowing dfs i'll know
Hey this must be very late and I have implemented this in my own java code , what would the time complexity be for this ? I was thinking O(n log(n)) ? But im not sure
I think there is 1 small bug in the code I see which went unnoticed. The key for border_islands should not just be 'ij' this would not work with double digits. The key should probably have a separator in between i and j. Also, this should be caught by the test cases. Please add more tests Clement.
Islands = copy of input loop on each values in the border of variable islands if value == 1 set it to 0, then recurse on each 4 directions in the recurse, if the value is 1 set it to 0 and recurse 4 directions at the end return intput - islands no need to store and query a separate visited structure
Below is my solution in js. I don't see anyone suggesting a DP solution so wanted to see if I'm thinking about the problem correct/am missing something. I solved this using DP. 1. One pass from top left to bottom right to mark nodes as "connected" based on whether top/left neighbors are borders/connected 2. One pass from bottom right to top left to mark nodes as "connected" based on whether bottom/right neighbors are borders/connected 3. Nodes marked as "not connected" after both passes are changed to 0 This is O(N*M) time and O(N*M) space but can be reduced to something like O(M) by only storing relevant rows for DP. Code is below: function removeIslands(map){ const isConnected = [] for(let i=0; i= map.length-1) return map[row][col] === 1 if(col === 0 || col >= map[0].length-1) return map[row][col] === 1 // now determine if neighbors are connected if(isConnected[row-1][col]) return true if(isConnected[row+1][col]) return true if(isConnected[row][col-1]) return true if(isConnected[row][col+1]) return true return false }
Actually u can do it easier with same the same time limit Run a dfs from each cell on the boarder that is 1 and you will mark all connected edges to that cell So basically u will mark all connected components that are connected to the boarder After that u know that each cell that is 1 and unmarked is part of an island (because it doesn't have a path to any boarder)
I somehow stumble upon this video and I really enjoyed it. Watching you two talk through the process was very insightful and it made it very easy to comprehend more so than if this was just a tutorial video.
My approach :- do bfs/ dfs from 1s present in first row, last row, first column and Last column and then change all the ones encountered to -1 After BFS/DFS Convert all 1s to 0 and -1 to 1
This interview is so co-operative and patient . The interviews which I have experienced are so brutal . They just kick you out once you answer a question wrong or lose interest and complete the interview for the sake of it . I hope I get an interview like theis , surely I will crack a good package ! Hoping for a better day ❤❤
I honestly can tell Clement enjoy these coding problems. These problems are like math problems to me. There’s multiple ways of doing these, but it’s always fun to find the most optimize solution. 😉
the way i would solve this is (i still didn't watch the vid) ... 1:mark all "1"s on the edges 2:go through the matrix from edge to center in a spiral or only horizontally (in c++ it would be for (int i =0; i< n^2) and mark every "1" that is connected to an already marked 1 and also mark every "1" that was connected to it but not marked 3:for loop to delete unmarked "1"s
I think there's a problem with the way he's naming the keys such that i = 1, j = 12 for example gives the same key as i = 11, j = 2 when they are two completely different coordinates. I think adding a space in between the i and j would solve this immediately though.
Did the test cases include matrices with more than 10 rows/columns? I expected the way Ben named the key to cause bugs. For example key "123" is ambiguous as (12, 3) and (1, 23) both map to this key - and that would mess up the visited map logic.
Great catch! As written, Ben's algorithm would indeed fail for some large matrices. We'd have to add a delimiter (like "-" or "|") between the indices in the key to correctly handle this. I've added a test case on AlgoExpert that would have failed Ben's algorithm!
I am posting this solution I thought of before watching the video so I do not know what Ben came up with. My idea is to remove the outer ones that are connected to the edges of the matrix and then subtract the obtained matrix to the original one. To do that, it’s actually not that bad if you’re familiar with percolation problems. Here is the idea : 1. put all the coordinates of the ones that are on the edges in a stack 2. While the stack is not empty : - take the top coordinate out of it (it is removed) - replace 1 by 0 - look at neighbours, if they are ones, add their coordinates in the stack (so at the top, this is important, otherwise you could add two identical coordinates and it would mess up with the algorithm) To make sure you don’t put two identical coordinates, we can add the condition that it is not possible to add a boarder coordinate to the stack. Thanks for this video, I got once again to find about a nice algorithmic problem :)
Actually, while going over the neighbours, changing the ones to zero and storing their coordinates would prove to be a lot more efficient. Same when storing the egdes coordinates in the stack
start from a blank matrix then start a recursive "walking" starting from the boders on the original matrix in all directions (recurssivly) on 1's and if it is valid put 1's on the new matrix (y and y coordinates must be maintained on the recursive process )
There is another optimization, you can skip checking every 2nd square. Like a checker board. Because if each square checks the 4 neighbors, then the next neighbor doesn't need to check any neighbors, because the neighbors are already going to check that for it. So if there were all 0's on the board, you could cut the time in half, less so if more 1's, however that's already optimized by skipping islands.
Just my approach to it would be. For (1 to less than length-1){ For(1 to less than length of inside array -1){ Check side ways neighbour and up and down neighbour by -1 +1 on index if it's 1, If all checks prove its island, replace it with 0 } }
This has a bug if the dimensions of the matrix have 2 digits. In your border_island dictionaries when you store Row 1, Col 12 and Row 11, Col 2 both of these are going to be represented as {'112', True } which are both same ... you probably should have used tuples to store them ... But I love this.
1. do a bfs(x,y) if current (x,y) is equal to 1 2. if up, right, left, down index is zero and up, right, left, down is in side bound then set the current index as 0
Thank you for the vulnerability. Good job! For a mid-level position, I might be a bit wary after this particular performance. Although we got there in the end, we needed a lot of hints/redirection from the interviewer and the communication could use some improvement. Clément mentioned the big issues. Some other things to be careful about: Jumping into the code too quickly before explaining one's thought process. This can help prevent excessive coding by coincidence as well.
There was a potential issue with the key. The key for the border_islands dictionary was just concatenation of the row & col indices. So, the index (1,21) and index (12,1) for instance will have the same key "121", which may mess up the visited checking for some different test cases.
Great catch! As written, Ben's algorithm would indeed fail for some large matrices. We'd have to add a delimiter (like "-" or "|") between the indices in the key to correctly handle this. I've added a test case on AlgoExpert that would have failed Ben's algorithm!
Great interview! Really enjoyed this question. I have one question about the code -- shouldn't there be a comma in the key? Like "0,0" instead of "00"? The way the code stands, the key could refer to multiple different coordinates. For example, "2020" could refer to both positions (20,20) and (202,0).
Yep, I was thinking about the same thing as soon as he wrote that. The sizes of the matrix in the test cases must have been small enough that there were no such cases. We will need it as soon as we enter two digits. 110 -> 1,10 or 11,0
@@СергейБарсуков-н8у tuple is an option or just add a dot in between {}{}. "{}.{}".format(i,j). Is there any advantage for string as a key or tuple in terms of hash collision?
I have found that if a problem requires checking border is reachable from a node in matrix, it is usually easier to invert the problem. Aka what nodes are reachable from the 4 borders. So you could have constructed a list of nodes that are touching edges and skip them in a second pass turning everything else to zero.
Clement I haven't seen the full video but the first problem was literally easy medium-ish. For boundary elements you can just do DFS/BFS and mark the connected elements as connected to the boundary of matrix and then finally iterate the matrix and remove the 1s not connected with the boundary 1 elements. This would be O(rows * columns) time and space complexity. We might optimise the space further.
These matrix problems are the ones that instantly make me nervous in an interview. "reverse rows or cols so the top left quad has the maximum sum... dead"
Interesting! However there is a bug in border_island_key function if one of the dimensions of the matrix exceeds 10. The key is ambiguous for (1, 11) & (11, 1) for instance (which both give '111'). Just inserting a space between both indexes will solve the issue.
A Card has 4 attributes (color, count, shading, shape), each attribute can have value 0, 1 or 2. 3 cards are considered as a set if for each attribute, 3 cards either have the same value or have different value from each other. For example: card1 (2, 0, 1, 2), card2 (1, 0, 0, 1) and card3(0, 0, 2, 0) are 1 set. card1(2, 0, 1, 2), card2 (1, 1, 0, 1) and card3(0, 1, 2, 0) are not 1 set. write a boolean function with 3 cards as input. This function returns true if they are 1 set, otherwise returns false. Now given a collection of cards, return true if there is a set of cards exsits, otherwise return false.
Nice and easy if you map it to a bitboard and can use bitwise operators to find your borders and store your islands as ints. Also makes it very fast on large matrices :)
for the string formatting thing (using '{}{}'.format(i, j) as a dict key): you can use tuple (i, j) as the dict key as tuples are hashable. No need for a key generator function or that fancy stuff also for your rec() function. you can create a global BORDER_ISLANDS dict and pass this in as a default argument: BORDER_ISLANDS = dict(). # global object def rec(i, j, border_islands=BORDER_ISLANDS): # default arg is this global object and so all recursive calls use the same object you wont need to pass this object around and all functions will use the same single instance of BORDER_ISLANDS.
Immediately after seeing him make the key by concatenating string representations of integers, I knew there would be an issue but sadly the test cases didn't cover it somehow. But basically, if *i* = 1, *j* = 11, the key generated for that is identical to the key generated when *i* = 11, *j* = 1. So basically he has no way to differentiate them, right? steps to generate test case his code will fail: 1. create a 13x13 matrix of zeroes 2. set matrix[1][11]=1 3. set matrix[11][1]=1 4. set matrix[12][1]=1 to connect the previous to the border since the step 3 tile is connected to the border, the key "111" is in the *border_islands* map, and this will prevent the tile at *matrix[1][11]* from properly being set to 0 as it should. if my logic or reasoning is wrong, let me know.
I noticed this right away. Actually I expected Ben would perform much better than this. If I were the interviewer, my opinion on the aspect of the algorithm would be no/weak hire. But he communicates so well, hence overall I'd give a weak hire / hire.
I’m just quickly throwing together this solution, so it might not be 100% right, but here we go: - create a recursive function and call it on every cell - return false if current cell is a 0 - return true if cell is on the edge (we know it is 1 at this point so don’t need to check) - set current cell to 0 temporarily - recurse over all 4 cardinal directions and store if any of them return true - set current cell to 1 if the stored value is true, 0 if false - return the stored value I didn’t code this at all and didn’t spend too much time thinking about this, so it might be a bit flawed. Feel free to let me know if it is and I will attempt to correct it.
Hahaha Clement was being nice to Ben when he said he would be a lean hire and it went to Ben's head so Clement had to stop being nice and just be honest with him 😂
Before watching the video tried to make it. I spent, maybe, half an hour on this. I made a copying of the matrix ('result' matrix). with values -1, 0, 1. -1 -> unknown, 0 -> no land, 1 -> land. Edges are copied and the middle is filled with -1. in "while(true)" it goes through every -1 value and checks if there is value 1 on the original matrix at its location and near it on the result matrix then it puts 1 on the result. When the cycle can't put 1's anymore, it breaks from the cycle and replaces all -1's with 0's on the result. There were faster variants, but this one was simple
@@nag0074 Task1 was fine it was designing a website ,, task 2 they asked me to build a calculator with JS and HTML ,, I am actually a beginner in JS so my = button was almost "functionless" 😂 I wish I could build websites with python it is way better tbh 😂
Here would be my approach: Create a set. Walk the perimeter and call explore() function for each cell of the perimeter. The function would add the cell index to the set and call itself on all adjacent cells. If passed cell is 0 or is already in the set, the explore() function will just return. Once it is done, walk the matrix and convert all 1 that are not in set to 0.
I dont know why am i here, but i like it. I'm 30 and randomly started liking coding/software development and got reminded of C++ programs i made in school days😂 Can i make money in this field in a year or two if i start today? I m very much tempted
Obviously itd be better if you had a 4 year CS degree but if you self teach web dev or something and have some college education, its possible to find a job in this field and make pretty dam good money
Recursively check each index with value 1 the top, left, right, bottom for value of 1 until at border. Return true if 0 at border, false if 1 at border.
Bro that's super interesting that Ben awad does the same steps trick at 23:35 when dealing with 2d matrices as I do. When needing to check all directions (including diagonal), I usually make a couple for loops to make the directions for me so i dont even have to hard code it. Explanation: Lets say we want to check in all directions (up, down, left, right, upright, downleft, upleft, downright) When we want go up, row changes by -1, and col doesnt change. When we wanna go down, row changes by +1, and col doesnt change. When we wanna go left, row doesnt change but col goes down by 1. etc etc. So, in the format [rowchange, colchange], the ways each direction changes the coordinates is: up: [-1, 0] down: [1, 0] left: [0, -1] right: [0, 1] upright: [-1, 1] downleft: [1, -1] upleft: [-1, -1] downright: [1, 1] This is actually every pair combination of -1, 0, and 1, not including [0, 0] (obviously because 0,0 is no change at all) So instead of hardcoding this, you can just make the combinations in python like so: for rChange in range(-1, 2): for cChange in range(-1, 2): if not (rChange is 0 and cChange is 0): new_row_coord = row_coord + rChange new_col_coord = col_coord + cChange #now do whatever u want with the new coordinates
get the position of all 1s . for each 1 check the elements at the index(+1 ,-1 , +length, -length(length of row)) till we reach the max length possible and store if it is connected to island. Do this for each 1's and we can exit if we hit 1 which has connected to island flag and remove the 1's which are not in boarder which are position of multiples of row length.
There is still a bug in the final solution. The keys used for the dictionary can collide. For example, positions row = 11, col = 1 and row = 1, col = 11 (and many others) are mapped to the same string '111'. That would cause weird results and would be hard to debug. Using a template like '{},{}' or using a tuple (row, col) as key would solve the issue
Could also create a string key with a separator like row + " : " + column. This string is easily hashable. Some languages make hashing collections difficult and this may be the best option.
1. Do DFS starting at all the 1's along the border, set them all to 0's. Be sure to use a visited matrix to avoid duplicate work. 2. Compare every cell of original matrix with matrix from (1), call it X 3. if origina[i][j] == 1 and X[i][j] == 1 then set original[i][j] = 0. All border connected islands will be 0 from (1) so whatever islands remain are the non-border connected islands. 4. return orignal
I’m currently learning JavaScript and html/css real quick. Picking up some best practices in general (code for scalability and multi-application use, and keep it organized. Also learning that extensions are also best practice (really fun 😂) and help speed up the coding/learning processes/efficiency process. Also I think green text (don’t know what it’s called yet) is good for organizing your code and if you ever need to come back to it later.
I'd use a bfs by putting all coordinates on the edge that have value 1 into a queue. Then work off the queue, mark each discovered cell (eg. by setting the value to 2), and add neighboring cells with value 1 to the queue. Then just iterate over the matrix and set every 1 to 0, and every 2 to 1. Time complexity O(n * m), extra space complexity O(1).
Just throwing an idea out there: For each row, I think I'd identify each one that matches "010" because that means it has no horizontal neighbors. I'd keep it's x,y coords in a horizontal_ones dict. Then, for each column, I'd do the same thing, looking for the same "010" pattern, those of course go into the vertical_ones dict. The overlap, that is, the elements present in both the horizontal_ones and vertical_ones dicts would be islands and be removed. The center part is done now. Now, for the edges, I think I just realized that if you took this grid, maybe it's height and width are 10, if you just put the whole thing in the center of a 12x12 grid of 0s, then you can use the same logic to do the edges. That's not very memory efficient, but it's definitely lazy efficient, because I wouldn't have to write any more logic to deal with the edges.
I value the thought process. The question about how to make sense of the 1 in the second row literally is a eureka moment that his code isnt gonna work. Because he still hasnt done the inner most square. The follow up question was litterally a saving grace lol
Basically We have to go through all the boundary elements and look for the 1 connected with it and change them to 2. Later we will check if element is 2 change it to 1 and element is 1 change it to 0.
I suck at coding interviews. Have done a few of them and it's just getting too into your head (focusing on stuff that isn't important yet) and wanting to just solve it on your own instead of vocalizing it. Something I struggle with and likely will not get over cause it's a personality thing. But that's fine cause I'm okay with my hybrid Design/UX Engineer skills instead of going full on frontend, tbh.
So, should we do a hard Google coding interview next? 👀
Be sure to check out the other video that we filmed on Ben’s channel, where he gave me an intermediate React interview! th-cam.com/video/6s0OVdoo4Q4/w-d-xo.html
PLEASE DO!!!!
sure 👀
Yes
Absolutely! Por favor!
Yes.
I should stick to inverting binary trees
Time to go back to your root(s)
Thank you for you positive, it was very pleased to watch this video.
Lmao
You did ok
Yesterday I was taking Facebook screening interview and they want you to solve this FAST
There’s something about seeing people being nervous for a mock interview like this that’s so satisfying
😂Loving to witness other people's misery, huh?
@@clem Yeah, I just hope my future interviewers are not seeing this comment 🤫
coding while the whole world is watching your competency and you are in position to prove it. Nah I would rather prefer real interviews.
Sadist😂😂
A villain in the making😂
Always love how Ben Awad deduces his problems. 50% Troll 50% Serious
"You know what? I'm being a pleb I think."
He was over-engineering this problem completely.
Donald Duck Coding, is amazing.
Watched this the day after you released Clement. During an interview with a fintech startup a week later, they gave me this exact problem. I was able to remember and successfully work through roughly 80% of it. Got offered the job and ultimately rejected their offer because one interviewer who would have been my daily collaborator, was rude and demeaning. All in all, this is great information and a great video you've produced here!
Amazing story! Thanks for sharing!
how much was the offer you declined? out of curiosity
Nice! Fintech companies IME can be horrible places to work because many are fundamentally banks/insurance companies at heart with tech departments. You'll be forced to appease some pretty unpleasant people in Fintech and try as they might Fintech just can't get the whole "tech" vibe good companies have.
Sometimes you just got to man up and work with A-holes.
@@Rob-vg6lw I come from a construction and ranching background; worked with plenty of A-holes. Working with A-holes doesn't make you a man. Ethics, love, and principles do. Thankfully I have those, and am now in a much richer and better environment and am in control of my future.
Word of advice - whoever told you it's manly to take a miserable job over better options, was incredibly wrong. Sometimes you may have no other option, but when you can afford to take a different job with a more supportive group, do it.
Me after 5 minutes: I don't think I am made for this
Trust me. I used to feel the same. Just keep at it, doing harder and harder probs and then you're solving such problems in no time :P
*le Clement after seeing this
"...and how to make sure that never happens to you,,, ALGOEXPERT"
@@anonymousasdoasidjasd9911 Best comment. lol
Don't worry the guy getting interviewed isn't either.
@@gujeffrey5433 For real. It's hard to stay motivated when interview coding questions are so contrived.
I think this shows that you need to practice specifically for these coding interviews. It doesn't matter if you are a programmer or you can code or if you are smart and you have experience. Ben is all these things. You need to practice problems like these to get a job at a company that does these interviews. It's like preparing for an exam. Even if you do ok like Ben did someone will do better because he spent hours and hours practicing. So many people want to get in these companies that it is very competitive. It's just how it is
^^^THIS. I am nowhere as knowledgable as Ben in technologies that are actually useful on the job, but because I have done many similar problems, it only took me 10 min to code up the solution. I didn't even have to think, my hands just spit out code from muscle memory lol
Same, came up with a solution within the first 3-5mins because I've been doing these non stop, everyday for the last 4-5 weeks
@@ducthinh2412 yup as soon as he showed the question, I knew this was dfs lol
@@ducthinh2412 where you do problems like these ? Gfg, leetcode etc ?
I'm 17 and I almost always have a solution to leetclde medium or easy level problems in less than a minute. Hard usually needs some testing so that the solution is fast enough.
Damn, this is a great coding interview question! ;)
I wonder who filmed the video explanation for it on AlgoExpert 👀
@@clem 👀
Almost two months ago, I started this video but didn't finish it because I wanted to solve this problem for myself. I didn't find a way to do it on my own, and both the problem and the video sat in the back of my mind for weeks. Today, after quite a lot of reading and practice (on other problems!), I returned to the problem with a fresh point of view and solved it in an hour (plus a few minutes), in one sitting, without any outside reference. It's been tough measuring my progress over the last few months, but thanks for giving me a point of reference!
You'd think that all software developers are building search engines. Lmfao. These coding questions can be solved, but the majority of problems in the software industry revolve around devops and system maintenance. Sometimes building new extensions / microservices / libraries, but these kind of problems only present themselves in my own pet (hobby) projects. Basically you can solve all of these abstract problems and still be a noob by lack of experience. Imo, time is much better spent by studying apis.
@@kimgysen10 dont forget actualy coding something thats where you notice a lot of problems and learn how important it is to write clean code
@@frankzander6234 For hobby projects yes. Majority of jobs as developer in the corporate world to earn a living requires quite a different skillset. That is in my experience.
@@frankzander6234 right but i'm not sure youre getting his point
@@kimgysen10 I don't think he's trying to progress professionally when he wrote this comment, the progress he's talking about is the progress he's made in how he thinks and his ability to problem solve. The relevancy of the project is none.
Here's my approach
1. Loop all edges - Only edges
2. If it's 0, then move to next
3. If it's 1, then update to 0 and find adjacent 1 and update to 0 until it can't find adjacent 1
At the end of this loop, you end up with a new matrix where 1s connected to edge is all 0s
then finally, you subtract the new matrix from the original.
Since all edges and connected cells are turn to 0, you're not changing anything, but the island cells will be 0
Nice!
That's a smart and out of the box approach.
thats rlly cool approach
So, dfs
I came up with this too, but went a step further and stored all of the 1's found on the edge in a stack to dfs over all in one go
I like how both of them interview each other and post the video on same time
Seeing this level of skill, almost feel as if I am not worth my salt to be in this industry. As someone who does app development for a living, this was overwhelming and quite literally humbling to watch.
seriously? his thinking was really messy。i had watched some college kid wrote a much more complicated board game in one sit including all the gui code in c without internet access，and almost no bugs. This guy was like without thinking and just try
That was not impressive at all I don't think. It felt to me like early monday morning level of coding, or really nervous level of coding. Maybe if it was me in his place I would do stupid stuff too, but either way, he didn't impress me :)
@@razrgu3838 mildly ADHD maybe (i assume)
I'm an experienced software developer but I would have definitively not passed that test. These Google interviews are all about algorithms which is something I learned twenty years ago in my studies but in my professional life I hardly ever came across algorithms again. I think this kind of interviews are pretty artificial and to pass them you have to train for it. I'm pretty sure it won't have much to do with your real work you will do.
@@rolfspuler1056 in which company do you work?
for the "is_border" and "is_border_matrix" functions, since returning True and False over an If statement is redundant, I would have simply done a one-liner: "return ". That would tell me he's been through a number of code reviews or is part of a larger team.
I've done my share of interviewing candidates as a Senior Eng at a huge retail company and even for me, a Google interview is anxiety-inducing.
Well done Ben! Coding when the entire world is watching - it's a nightmare. You're a really brave guy!
Ben laughing when Clement stops him for a second is great cause he knows it's a jank approach.
Please keep doing these! This is great and love hearing the thought process
Mark all the ones you shouldn’t touch with a different value. You can use DFS to do this.
Go over the entire grid at the end, changing the marked cells back to one, the 1s to zero, and the 0s skip.
As someone else pointed out, BFS would have time complexity O(min(N, M)) whereas DFS would have time complexity O(N*M)
I'm basically new to code. I am currently learning the very basics of C++. This video is way harder to understand for me but i watched either way. I will rewatch it again in a few years and hopefully my skills are better and i can be able to understand it. Great content!!
Don't give up
This is your 2 week reminder to keep practicing
Adding on to Ryan’s comment. How is your understanding of the video now? I’m about 75% of the way through Codecademy’s Full Stack development course and I can understand the majority of what’s going on, but still am lost in some places.
@@icyrex7893 do you understand it all now?
As a computer scientist with ten years in industry, it was neat but nothing crazy: how is your understanding now?
I'm addicted to these mock interview. Thank you for these quality content.
This wasn't easy. Good job Ben. Good job coach Clément
Clément seems like a really good interviewer which is hugely helpful.
The "if return True else return False" lines are killing me :)
After getting harassed by teachers all year when someone did this, it also triggers me ^^
@@NirEndal how are you supposed to do it?
@@ButerWarrior44 "return something" with an eventual cast to boolean if it is not a boolean already
@@NirEndal pseudocode please
@@ButerWarrior44
Instead of:
IF condition
THEN RETURN TRUE
ELSE
RETURN FALSE
Use:
RETURN condition
I can't do anything more explicit ^^'
thank you for making these videos! I solved this problem by modifying the grid to have perimeter of 1s (so both len(grid) and len(grid[0]) increase by 2) and running BFS on the grid. After finding an island, all other islands are changed to 0s. Pausing the video after you explain the question and talking out loud "back" to you has been so so helpful.
This is my favorite crossover
Looking forward to solving hard coding challenge with Ben... that would be fun for sure lol
really love this format! so entertaining! very interesting to see "live" how ppl approach a problem
This is extremely inspiring some how… Well done.
I absolutely love these videos! Do more of these interviews !
Simple and straightforward - DFS from each position on the border if its marked as 1 and mark everything in the dfs as 2. Then remove all 1 from the matrix and convert all 2 back into 1. O(n) time O(1) space.
Just wanted to say a nice very fast solution I thought up to this problem:
1) Make a second matrix (we'll call it m2) of the same size filled with zeroes
2) Iterate over the edges of the original matrix.
3) Each time you encounter a 1, do a depth first search. At each visited position, add a 1 to the corresponding position in m2 and then set the value to 0 in the original matrix.
4) Once you've iterated all edges, m2 will contain all islands that connected to the edges, though none that didn't. Return m2.
why do you have to set the value to zero in the original matrix
@@Lorkwondo1234 just helpful to avoid having to keep track of which positions you've visited. Setting to 0 means future function calls that check a cell which has already been visited just return.
Oof i don't know what is dfs yet, will learn about it and come back 🤠
I was thinking the same approach but didn't knew how to determine if it's on the edge or not. Maybe after knowing dfs i'll know
Hey this must be very late and I have implemented this in my own java code , what would the time complexity be for this ? I was thinking O(n log(n)) ? But im not sure
I guess one downside with this is additional space complexity needed for an extra matrix
We waited a lengthy period for this, but it was worth it🙌
I think there is 1 small bug in the code I see which went unnoticed. The key for border_islands should not just be 'ij' this would not work with double digits. The key should probably have a separator in between i and j. Also, this should be caught by the test cases. Please add more tests Clement.
I was surprised when the host didn't noticed that
Another solution would be to add the key as a tuple
key[(1,2)] = True
Islands = copy of input
loop on each values in the border of variable islands
if value == 1 set it to 0, then recurse on each 4 directions
in the recurse, if the value is 1 set it to 0 and recurse 4 directions
at the end return intput - islands
no need to store and query a separate visited structure
Below is my solution in js. I don't see anyone suggesting a DP solution so wanted to see if I'm thinking about the problem correct/am missing something.
I solved this using DP.
1. One pass from top left to bottom right to mark nodes as "connected" based on whether top/left neighbors are borders/connected
2. One pass from bottom right to top left to mark nodes as "connected" based on whether bottom/right neighbors are borders/connected
3. Nodes marked as "not connected" after both passes are changed to 0
This is O(N*M) time and O(N*M) space but can be reduced to something like O(M) by only storing relevant rows for DP.
Code is below:
function removeIslands(map){
const isConnected = []
for(let i=0; i= map.length-1) return map[row][col] === 1
if(col === 0 || col >= map[0].length-1) return map[row][col] === 1
// now determine if neighbors are connected
if(isConnected[row-1][col]) return true
if(isConnected[row+1][col]) return true
if(isConnected[row][col-1]) return true
if(isConnected[row][col+1]) return true
return false
}
Actually u can do it easier with same the same time limit
Run a dfs from each cell on the boarder that is 1 and you will mark all connected edges to that cell
So basically u will mark all connected components that are connected to the boarder
After that u know that each cell that is 1 and unmarked is part of an island (because it doesn't have a path to any boarder)
I somehow stumble upon this video and I really enjoyed it. Watching you two talk through the process was very insightful and it made it very easy to comprehend more so than if this was just a tutorial video.
My approach :- do bfs/ dfs from 1s present in first row, last row, first column and Last column and then change all the ones encountered to -1
After BFS/DFS
Convert all 1s to 0 and -1 to 1
Same
show your source code to check if it works
This interview is so co-operative and patient . The interviews which I have experienced are so brutal . They just kick you out once you answer a question wrong or lose interest and complete the interview for the sake of it . I hope I get an interview like theis , surely I will crack a good package ! Hoping for a better day ❤❤
as an ex-Googler, you must interview him in Angular >: )
I think he already did
Make a hard coding interview one! These two were so good!
I honestly can tell Clement enjoy these coding problems. These problems are like math problems to me. There’s multiple ways of doing these, but it’s always fun to find the most optimize solution. 😉
FYI Clement that Niagara Fall screen saver is nice. 🙂
the way i would solve this is (i still didn't watch the vid) ...
1:mark all "1"s on the edges
2:go through the matrix from edge to center in a spiral or only horizontally (in c++ it would be for (int i =0; i< n^2) and mark every "1" that is connected to an already marked 1 and also mark every "1" that was connected to it but not marked
3:for loop to delete unmarked "1"s
I love how Ben starts randomly laughing without any apparent reason
Clement be like
"I don't mean to interrupt you between your thought process, but, ANYWAYS Imma just interrupt you" Lmao, Clement is the best
You guys are just incredible and an inspiration, that helps me keep on learning. Thanks for sharing!!
I think there's a problem with the way he's naming the keys such that i = 1, j = 12 for example gives the same key as i = 11, j = 2 when they are two completely different coordinates. I think adding a space in between the i and j would solve this immediately though.
Wow Clements product has come a long way. So awesome!! Keep up the amazing work buddy.
This guy is an incredible programmer and his mind is a logic engine.
@Anna Meredith he meant these lol
I liked the question. Well done, Ben! and Clement, you were very helpful :)
Did the test cases include matrices with more than 10 rows/columns? I expected the way Ben named the key to cause bugs. For example key "123" is ambiguous as (12, 3) and (1, 23) both map to this key - and that would mess up the visited map logic.
exactly what I thought, he should have used tuples as keys
Yep, there must be a separator, like {}:{}
Or just use the correct datatype which would be a set. Then you don't have to assign it a value that is never used.
Great catch! As written, Ben's algorithm would indeed fail for some large matrices. We'd have to add a delimiter (like "-" or "|") between the indices in the key to correctly handle this.
I've added a test case on AlgoExpert that would have failed Ben's algorithm!
@@clem Did you just call Ben a failure ? 😂
I am posting this solution I thought of before watching the video so I do not know what Ben came up with.
My idea is to remove the outer ones that are connected to the edges of the matrix and then subtract the obtained matrix to the original one.
To do that, it’s actually not that bad if you’re familiar with percolation problems. Here is the idea :
1. put all the coordinates of the ones that are on the edges in a stack
2. While the stack is not empty :
- take the top coordinate out of it (it is removed)
- replace 1 by 0
- look at neighbours, if they are ones, add their coordinates in the stack (so at the top, this is important, otherwise you could add two identical coordinates and it would mess up with the algorithm)
To make sure you don’t put two identical coordinates, we can add the condition that it is not possible to add a boarder coordinate to the stack.
Thanks for this video, I got once again to find about a nice algorithmic problem :)
Actually, while going over the neighbours, changing the ones to zero and storing their coordinates would prove to be a lot more efficient. Same when storing the egdes coordinates in the stack
I could watch this 2 forever lol, amazing, entertaining and learned a lot :)
start from a blank matrix then start a recursive "walking" starting from the boders on the original matrix in all directions (recurssivly) on 1's and if it is valid put 1's on the new matrix (y and y coordinates must be maintained on the recursive process )
“I really hope I don’t refactor and just ruin my code even more” - that always happens 😂 34:16
There is another optimization, you can skip checking every 2nd square. Like a checker board. Because if each square checks the 4 neighbors, then the next neighbor doesn't need to check any neighbors, because the neighbors are already going to check that for it. So if there were all 0's on the board, you could cut the time in half, less so if more 1's, however that's already optimized by skipping islands.
Saying I'm not gonna be asked to find max element in an array at my interview
Just my approach to it would be.
For (1 to less than length-1){
For(1 to less than length of inside array -1){
Check side ways neighbour and up and down neighbour by -1 +1 on index if it's 1,
If all checks prove its island, replace it with 0
}
}
This has a bug if the dimensions of the matrix have 2 digits. In your border_island dictionaries when you store Row 1, Col 12 and Row 11, Col 2 both of these are going to be represented as {'112', True } which are both same ... you probably should have used tuples to store them ... But I love this.
Seems like he jumps into coding immediately without without even explaining what he will do in a simple pseudo code.
0:55 That Ben's laugh as Clement promoted AlgoExpert 🤣🤣
ha h a
Great coding interview ! Brilliant stuff, Clément !
When you do functions with a single boolean expression, just do return boolean expression rather than if/else.
1. do a bfs(x,y) if current (x,y) is equal to 1
2. if up, right, left, down index is zero and up, right, left, down is in side bound then set the current index as 0
This is the reason i hate interviews like this. I could solve this for sure just might take a week
It will come to me while I’m washing my hair in the shower
and this is why you wont work in google
@@damf7106 true, but i already have 6 figure salary, so i am good.
@@archmad you work as a programmer?
Thank you for the vulnerability. Good job! For a mid-level position, I might be a bit wary after this particular performance. Although we got there in the end, we needed a lot of hints/redirection from the interviewer and the communication could use some improvement. Clément mentioned the big issues. Some other things to be careful about: Jumping into the code too quickly before explaining one's thought process. This can help prevent excessive coding by coincidence as well.
There was a potential issue with the key. The key for the border_islands dictionary was just concatenation of the row & col indices. So, the index (1,21) and index (12,1) for instance will have the same key "121", which may mess up the visited checking for some different test cases.
Great catch! As written, Ben's algorithm would indeed fail for some large matrices. We'd have to add a delimiter (like "-" or "|") between the indices in the key to correctly handle this.
I've added a test case on AlgoExpert that would have failed Ben's algorithm!
@@clem could also use a tuple of the indices as a key. my world was changed when I learned this was possible.
This was AWESOME. Thank you!
Great interview! Really enjoyed this question. I have one question about the code -- shouldn't there be a comma in the key? Like "0,0" instead of "00"? The way the code stands, the key could refer to multiple different coordinates. For example, "2020" could refer to both positions (20,20) and (202,0).
Yep, I was thinking about the same thing as soon as he wrote that. The sizes of the matrix in the test cases must have been small enough that there were no such cases. We will need it as soon as we enter two digits. 110 -> 1,10 or 11,0
I`s better to use a tuple as a key here
@@СергейБарсуков-н8у or a Point
@@СергейБарсуков-н8у tuple is an option or just add a dot in between {}{}. "{}.{}".format(i,j). Is there any advantage for string as a key or tuple in terms of hash collision?
I have found that if a problem requires checking border is reachable from a node in matrix, it is usually easier to invert the problem. Aka what nodes are reachable from the 4 borders.
So you could have constructed a list of nodes that are touching edges and skip them in a second pass turning everything else to zero.
It's funny how ben always just wants to start programming without fully explaining how his algorithm works
Astonishing interview question!
Really enjoyed it!
Clement I haven't seen the full video but the first problem was literally easy medium-ish. For boundary elements you can just do DFS/BFS and mark the connected elements as connected to the boundary of matrix and then finally iterate the matrix and remove the 1s not connected with the boundary 1 elements. This would be O(rows * columns) time and space complexity. We might optimise the space further.
Ok, do you want us to praise you now?
Love it xd
These matrix problems are the ones that instantly make me nervous in an interview. "reverse rows or cols so the top left quad has the maximum sum... dead"
Interesting! However there is a bug in border_island_key function if one of the dimensions of the matrix exceeds 10. The key is ambiguous for (1, 11) & (11, 1) for instance (which both give '111').
Just inserting a space between both indexes will solve the issue.
I was just going to write a comment about that too 😀
A Card has 4 attributes (color, count, shading, shape), each attribute can have value 0, 1 or 2. 3 cards are considered as a set if for each attribute, 3 cards either have the same value or have different value from each other. For example:
card1 (2, 0, 1, 2), card2 (1, 0, 0, 1) and card3(0, 0, 2, 0) are 1 set.
card1(2, 0, 1, 2), card2 (1, 1, 0, 1) and card3(0, 1, 2, 0) are not 1 set.
write a boolean function with 3 cards as input. This function returns true if they are 1 set, otherwise returns false.
Now given a collection of cards, return true if there is a set of cards exsits, otherwise return false.
Nice and easy if you map it to a bitboard and can use bitwise operators to find your borders and store your islands as ints. Also makes it very fast on large matrices :)
😂
for the string formatting thing (using '{}{}'.format(i, j) as a dict key):
you can use tuple (i, j) as the dict key as tuples are hashable. No need for a key generator function or that fancy stuff
also for your rec() function. you can create a global BORDER_ISLANDS dict and pass this in as a default argument:
BORDER_ISLANDS = dict(). # global object
def rec(i, j, border_islands=BORDER_ISLANDS): # default arg is this global object and so all recursive calls use the same object
you wont need to pass this object around and all functions will use the same single instance of BORDER_ISLANDS.
Immediately after seeing him make the key by concatenating string representations of integers, I knew there would be an issue but sadly the test cases didn't cover it somehow.
But basically, if *i* = 1, *j* = 11, the key generated for that is identical to the key generated when *i* = 11, *j* = 1. So basically he has no way to differentiate them, right?
steps to generate test case his code will fail:
1. create a 13x13 matrix of zeroes
2. set matrix[1][11]=1
3. set matrix[11][1]=1
4. set matrix[12][1]=1 to connect the previous to the border
since the step 3 tile is connected to the border, the key "111" is in the *border_islands* map, and this will prevent the tile at *matrix[1][11]* from properly being set to 0 as it should.
if my logic or reasoning is wrong, let me know.
I noticed this right away. Actually I expected Ben would perform much better than this. If I were the interviewer, my opinion on the aspect of the algorithm would be no/weak hire. But he communicates so well, hence overall I'd give a weak hire / hire.
I’m just quickly throwing together this solution, so it might not be 100% right, but here we go:
- create a recursive function and call it on every cell
- return false if current cell is a 0
- return true if cell is on the edge (we know it is 1 at this point so don’t need to check)
- set current cell to 0 temporarily
- recurse over all 4 cardinal directions and store if any of them return true
- set current cell to 1 if the stored value is true, 0 if false
- return the stored value
I didn’t code this at all and didn’t spend too much time thinking about this, so it might be a bit flawed. Feel free to let me know if it is and I will attempt to correct it.
Oh and I came up with this before a solution was coded (as in, I came up with this right after the problem was explained).
Hahaha Clement was being nice to Ben when he said he would be a lean hire and it went to Ben's head so Clement had to stop being nice and just be honest with him 😂
Before watching the video tried to make it. I spent, maybe, half an hour on this. I made a copying of the matrix ('result' matrix). with values -1, 0, 1. -1 -> unknown, 0 -> no land, 1 -> land. Edges are copied and the middle is filled with -1. in "while(true)" it goes through every -1 value and checks if there is value 1 on the original matrix at its location and near it on the result matrix then it puts 1 on the result. When the cycle can't put 1's anymore, it breaks from the cycle and replaces all -1's with 0's on the result. There were faster variants, but this one was simple
I have a coding competition tomorrow please wish me luck
What happened?
@@nag0074 Task1 was fine it was designing a website ,, task 2 they asked me to build a calculator with JS and HTML ,, I am actually a beginner in JS so my = button was almost "functionless" 😂 I wish I could build websites with python it is way better tbh 😂
Here would be my approach:
Create a set.
Walk the perimeter and call explore() function for each cell of the perimeter. The function would add the cell index to the set and call itself on all adjacent cells. If passed cell is 0 or is already in the set, the explore() function will just return.
Once it is done, walk the matrix and convert all 1 that are not in set to 0.
That is exactly what i did yesterday :-)
I dont know why am i here, but i like it.
I'm 30 and randomly started liking coding/software development and got reminded of C++ programs i made in school days😂
Can i make money in this field in a year or two if i start today? I m very much tempted
Obviously itd be better if you had a 4 year CS degree but if you self teach web dev or something and have some college education, its possible to find a job in this field and make pretty dam good money
Recursively check each index with value 1 the top, left, right, bottom for value of 1 until at border. Return true if 0 at border, false if 1 at border.
Bro that's super interesting that Ben awad does the same steps trick at 23:35 when dealing with 2d matrices as I do. When needing to check all directions (including diagonal), I usually make a couple for loops to make the directions for me so i dont even have to hard code it.
Explanation:
Lets say we want to check in all directions (up, down, left, right, upright, downleft, upleft, downright)
When we want go up, row changes by -1, and col doesnt change. When we wanna go down, row changes by +1, and col doesnt change. When we wanna go left, row doesnt change but col goes down by 1. etc etc.
So, in the format [rowchange, colchange], the ways each direction changes the coordinates is:
up: [-1, 0]
down: [1, 0]
left: [0, -1]
right: [0, 1]
upright: [-1, 1]
downleft: [1, -1]
upleft: [-1, -1]
downright: [1, 1]
This is actually every pair combination of -1, 0, and 1, not including [0, 0] (obviously because 0,0 is no change at all)
So instead of hardcoding this, you can just make the combinations in python like so:
for rChange in range(-1, 2):
for cChange in range(-1, 2):
if not (rChange is 0 and cChange is 0):
new_row_coord = row_coord + rChange
new_col_coord = col_coord + cChange
#now do whatever u want with the new coordinates
is there an official name to this method that I can look up in greater detail?
get the position of all 1s . for each 1 check the elements at the index(+1 ,-1 , +length, -length(length of row)) till we reach the max length possible and store if it is connected to island. Do this for each 1's and we can exit if we hit 1 which has connected to island flag and remove the 1's which are not in boarder which are position of multiples of row length.
Do a hard one with Ben Awad
that sounds so wrong
we'll allow him to solve in O(n!) or O(n^n)
I just solved this on AlgoExpert by myself and came back to this video to get a feel of how this question would've felt in the coding interview
Ben laughing at his ratings has me dying lol
There is still a bug in the final solution. The keys used for the dictionary can collide. For example, positions row = 11, col = 1 and row = 1, col = 11 (and many others) are mapped to the same string '111'. That would cause weird results and would be hard to debug. Using a template like '{},{}' or using a tuple (row, col) as key would solve the issue
Could also create a string key with a separator like row + " : " + column. This string is easily hashable. Some languages make hashing collections difficult and this may be the best option.
@@nonconsensualopinion It's what I proposed with the template '{},{}'. I used the comma as separator
I thought i was bad at problem solving until i saw ben who i thought was pretty good at it being very bad at it
1. Do DFS starting at all the 1's along the border, set them all to 0's. Be sure to use a visited matrix to avoid duplicate work.
2. Compare every cell of original matrix with matrix from (1), call it X
3. if origina[i][j] == 1 and X[i][j] == 1 then set original[i][j] = 0. All border connected islands will be 0 from (1) so whatever islands remain are the non-border connected islands.
4. return orignal
Question for Ben and Clément! Can y’all each make a video for best practices in React in front end and best practices in Python for algorithms?
Give up
I’m currently learning JavaScript and html/css real quick. Picking up some best practices in general (code for scalability and multi-application use, and keep it organized. Also learning that extensions are also best practice (really fun 😂) and help speed up the coding/learning processes/efficiency process. Also I think green text (don’t know what it’s called yet) is good for organizing your code and if you ever need to come back to it later.
I'd use a bfs by putting all coordinates on the edge that have value 1 into a queue. Then work off the queue, mark each discovered cell (eg. by setting the value to 2), and add neighboring cells with value 1 to the queue. Then just iterate over the matrix and set every 1 to 0, and every 2 to 1. Time complexity O(n * m), extra space complexity O(1).
Just throwing an idea out there:
For each row, I think I'd identify each one that matches "010" because that means it has no horizontal neighbors. I'd keep it's x,y coords in a horizontal_ones dict.
Then, for each column, I'd do the same thing, looking for the same "010" pattern, those of course go into the vertical_ones dict.
The overlap, that is, the elements present in both the horizontal_ones and vertical_ones dicts would be islands and be removed. The center part is done now.
Now, for the edges, I think I just realized that if you took this grid, maybe it's height and width are 10, if you just put the whole thing in the center of a 12x12 grid of 0s, then you can use the same logic to do the edges. That's not very memory efficient, but it's definitely lazy efficient, because I wouldn't have to write any more logic to deal with the edges.
It doesn't work with islands that are for example 2x2. Try for example: [[0, 0, 0, 0], [0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]
I value the thought process.
The question about how to make sense of the 1 in the second row literally is a eureka moment that his code isnt gonna work.
Because he still hasnt done the inner most square.
The follow up question was litterally a saving grace lol
Clement do you wanna be a Software Engineer at Google?? I just want a heart to type Megan's famous ad-line here 😂😂😂
Basically
We have to go through all the boundary elements and look for the 1 connected with it and change them to 2. Later we will check if element is 2 change it to 1 and element is 1 change it to 0.
I suck at coding interviews. Have done a few of them and it's just getting too into your head (focusing on stuff that isn't important yet) and wanting to just solve it on your own instead of vocalizing it. Something I struggle with and likely will not get over cause it's a personality thing.
But that's fine cause I'm okay with my hybrid Design/UX Engineer skills instead of going full on frontend, tbh.
Talking to yourself out loud is helpful