I'm a mid/senior level webdev and I just come back to Alvin's videos when preparing for interview everytime. its so simple and help me rebrush all of them without much hassle. very good work
I chose this video after I got an invitation for an online test from a well-known company. This was my only chance since I have never studied algorithm oriented programming. I was given a graph problem and handled it correctly with a minor deficiency in efficiency :) Thanks a lot!
This is the only course I’ve found that has successfully got the concept of graphs through my thick skull, it’s been a subject that’s been so hard for me to learn. Thank you for making this!!
This is phenomenal! I was finding Graph problems so difficult before watching this video, and now they seem fairly easy after watching the entire video! Thanks a ton!
How this person managed to explainamy concepts perfectly is superb... We really need people like this guy in the teaching field... I downloaded the video and I ve not regretted at all
This course is hands-down, unequivocally fantastic!!! Best useful course I've seen about graphs. It ties together the algorithms with concrete use cases which just clicked. Thank you for the fantastic job!!!!
This was amazing. I've watched a bunch of these types of videos over the years and this is the only one that actually made it look easy. I can actually say I understand this now. Thank you.
I never knew Graphs were that easy! You're one of the rarest gems I have come across in my life. Followed it all along. Solved all the problems in Python.
This is a great intro to basic graph problems - however we may need a part two to cover more advanced topics such as Union Find, Minimum Spanning Trees, Dijkstras & bellman ford, topological sort, etc.
For island count (1:58:20), you actually don't need a visited set to keep track of the tiles you have already visited. Simply set the land tile to water "W" when you visit it; you already have the logic there to skip over water tiles.
OMG this video is an absolute gem! I used to be baffled by the island problem and just simply memorizing the solution on leetcode, but after following along for two hours, I can confidently code up the solution all by my self! Thank you so much you are a fantastic teacher.
Before watching this course, I dreaded DFS BFS problems. I always used to somehow understand the logic just before the interview and then forget everything. But now, I understand the logic, visualization and how to appraoch it. Thank you so much Alvin and FCC for this. I am definitely going to watch other interview videos. 👍
Thank you Alvin! I've been trying to grind leetcode and DSA for interviews, and this video as well as your binary trees video was so amazingly helpful and way more efficient than blindly doing problems to help me strengthen the fundamentals, and in particular, I was really struggling with handling how to handle graphs in the form of a matrix rather than an adjacency list -- I typically tried to convert it into an AL unnecessarily, but watching this video, especially the numIslands section, made things just click. You also have given me so much of a better understanding and less fear of recursion, even going to appreciation of recursion! That's crazy. Also, for the numIslands problem, to save a bit on space, I got rid of the visited set and just set grid[row][col] == 'V' to mark that it's visited, and only add where grid[row][col] == L. This worked.
@@vinayaktyagi8773 Likely trees or linked lists next. You can check out my platform, Structy (link in the video description), if you want to check it out early before it premiers on free code camp.
This is an incredibly straight forward and simple to understand graph video. This video definitely made me clean up rough edges in my understanding of tackling graph DSA problems. Thank you !!
This is amazing. I found it really easy to understand the graph algorithms after referring to this video. Great job, Alvin :) Small correction in the solution described for checking if a source and destination have a path in an undirected graph. (line no. 8) The logic of traversing the neighbors and adding the source node to the visited list has to be part of the condition that validates whether ta node is visited or not. if !(visited.has(src)) { // add the src to the visitors list // recursively traverse through the src neighbors }
Just for correction, Islands count problem, inside DFS function for checking boundaries, for 'j' check for j < grid[0].length, because if in NxM grid N !== M, in that case j < grid.length cannot check for true boundary and result will be wrong, I just wanna mention that in case
I was confused when you started going through depth first search because you put both of A's neighbors B and C on the stack which seems to be breadth first. But what helped me to understand is the fact that the loop starts when we pop a node from the stack, not when we add it to the stack. The stack doesn't represent which node we are currently "at", the node we pop from the stack does that. The stack is just an in between step. So we don't actually "go to" C at first, we just add it to the stack for now. We do "go to" B because we pop it from the stack
Your explanation is ideal. Voice and the way you expalin does not make a listener tired. This is just amazing. Please continue doing your job. It is amazing
I am actually studying C but I have found this video tremendeously helpfull in explaining how graphs are traversed and how not to use return in a bad way when you want to actually make use of recursion to backpedal from dead ends in situations where you don't find your base case. thank you again. subscribed!
This was an elite video. I did the last two questions and did not understand how to approach it or understood what the right answer for those questions. Watch the 2+ hours of video as well as coding it up myself, and everything makes sense. EVERYTHING. I am now able to do one of the toughest topics that I was having trouble with. So BIG Thank You for that. P.S. For the last 2 questions, you don't need the visited set, you can flip the land to water or another value altogether. Avoiding the cyclic infinite loop.
Thanks Alwin for this great tutorial. I had been trying to understand and get an answer to a similar set of problems for a very long time. You made my day, ...probably my life!
This is really really a complete tutorial for “Graph based Problems” both in terms of coding Problems, interview preperation including MAANG and even for a new bie.
Here are the leetcode equivalent questions: has path/ undirected path - 1971. Find if Path Exists in Graph connected components - 323. Number of Connected Components in an Undirected Graph island count - 200. Number of Islands minIsland - 695. Max Area of Island
As an embedded software engineer, I never learned about these kind of algorithms. This is invaluable knowledge for me, thanks for leveling up my skill!
easily the best teacher of algorithms I've ever seen. Alvin spends time going over theory, Big O notation, and common errors. As someone that's never studied algorithms and data structures, all the videos he has done have helped drastically
34:29 for anyone confused here, the graph in the example given is incorrect. number of edges is not n^2. To show a graph with n^2 edges each node needs to point to every other node AND also point to itself
Thank you so much. This is the only video on UA-cam which addresses the graph algorithms the way they should be addressed - graphical visualization, pseudocode and actual problem solving.
great video, but i have a minor comment at 1:00:34, look at the hasPath function, i would swab the first 2 lines (line 7 and 8), which represent our 2 atomic cases, this way you can use the function to check whether there is a path from src to dst that does not pass through node x, or nodes {x, y, z} by calling the function with the visited set already populated with the nodes you want to exclude, this would not work with the current implementation but it still works fine
This is fantastic. Makes graphs easy to digest. Thanks for that. I did find a minor issue in the island count problem solution. It assumes that each row will have the same number of columns. That assumption isn't described in the problem. So a quick fix would be to run the nested loop to the length of the row instead of fixing it to zero. And doing the same for the inbounds check. That would present an issue in the exploration in that there might not be any map in either direction, so you'll have to add an existence check in the explore to account for that. This way if a row has less coordinates, or more, it would still work properly.
Preprocessing the input by storing edges in hashmap: this line took all my fears of not even trying to attempt graph problems to making me feel I could do any graph problems. Thanks for this video ❤️
Just have to agree with others that this is one of if not the best course on graph problems out there. Like many, I was intimidated by graph problems early on but this video has made me so much more comfortable with them. I feel like I actually understand them at a deep level now.
Great tutorial Alvin. For python programmers, here is the pattern for largest components (connected components is similar): def explore(arr, node, visited, comp_count): if node in visited: return 0, visited visited[node] = 1 for neighbour in arr[node]: comp_count += 1 explore(arr, neighbour, visited, comp_count) return comp_count, visited def largestcomponent(arr): largest_comp = 0 visited = {} for node in arr: if node not in visited: comp_count, visited_new = explore(arr, node, visited, 1) visited = visited_new if comp_count > largest_comp: largest_comp = comp_count return largest_comp
Thanks to the first 30mins of this video alone I was able to extend the idea of adjacency lists and independently solve all the following problems on undirected graphs and grid graphs in just under a week's worth of practice
Quick observation on the Nondirected - Has Path problem: Before that, Alvin, you are a genius man! You make everything sooo easy B) Ok here it goes: At around 59:00 you are using a set to keep track of visited nodes and returning false as soon as you find the source element in the set. I feel you are missing an edge case here. Imagine a graph in the shape of a Kite 🪁🪁🪁🪁 --> A cycle on top and a tail just below the cycle. The cycle and the tail are one single graph. Let us say the cycle starts at node a (a - b - c - a) and the tail (a - m - n - o). I want to find a path between (b and o) We need to to track the remaining nodes of the graph, because, without it, the process will start from a and take a trip through the cycle and come back to a; this already exists in set at this point in execution, the function returns false without checking the tail section of the Kite Shaped Graph.
I thought I'd seen the all the greatest of online coding instructors, and then I found Alvin... Seriously these explanations are paced exactly perfectly and ultra clear
One issue in the shortest path, 1:36:16 , its that whenever you find the destination you always return it. Its not always the shortest path. You can maintain a min counter to maintain the shortest distance. Also thanks for the wonderful video, just by following the videos , I could do the implementation on my own
I think that would have been an issue if we were using DFS, considering the nature of BFS and visited set that we are using, whenever we find the destination, that path will always be the shortest.
@@siddhantgupta4295 correct, thanks for the detailed explaination. I came to know about this after doing many bfs problems. I dont knw if it was mentioned in this video, if not then it should be mentioned :) happy coding
You just made me crack my Interview on graph algorithms... and I have just seen half the video only to make it... Thanks a ton Bro... Let me complete this
number islands problem couldnt have been solved easier than this, kudos to this guy! for once i was able to understand the solution and complete the problem without errors.
For the island count problem, i ended up converting the entire 2D grid into a graph containing only 'L' nodes with multiple components. I then did a dfs for each of those islands, incrementing a counter everytime i finished exploring one of them and adding it to my visited set. Really interesting how you can solve this problem in such a different way!
For largest component problem , there can be an easy and alternate solution . Since , we already have visited which is keeping track of all nodes. There is no need to maintain separate "size". Below is the code you can use to your advantage: const largestComponent = (graph) => { let maxSize = 0; let previousVisitedSize = 0; let actualSize = 0; const visited = new Set(); for(const node in graph) { if(explore(graph, node, visited) === true) { actualSize = visited.size - previousVisitedSize; maxSize = Math.max(actualSize, maxSize); previousVisitedSize = visited.size; } } return maxSize; }; const explore = (graph, node, visited) => { if(visited.has(String(node))) return false; visited.add(String(node)); for(const neighbor of graph[node]) { explore(graph, neighbor, visited) } return true; }
The series of your videos are amazing. They are the best courses I have ever seen regarding data structure and algorithm studies. However, I do think I find out a potential mistake. At 1:53:55, check the line 12, const colInbounds = 0
Thank you for the video. Also to find the minimum sizes of the island, we can use the method that calls the recursive function (parent method) to hold an integer that points to the size of the visited set before the recursive call and then use the difference in size after the recursive call to determine the minimum size. This to me didn't require me to create a new recursive method that returns an integer. However, creating an extra method does have its benefits too.
This guy is phenomenal. His course on dynamic programming was exceptionally well done as well.
A little old comment, but exactly my thought, I just went through his dynamic programming course and completely agree
@@mauricemarin5810 same here
Hi, @Chris can you add a link to the dynamic programing course. Thanks
@@elad7264 ua-cam.com/video/oBt53YbR9Kk/v-deo.html
Agreed
after that course, I am so confident on DP.
Einstein once said "If you are able to explain it to a 5 year old, you understand it yourself". Alvin, you are one of a kind!
Richard Fynman said that.
@@syedtalha1264 I also said that, hey!
Another immortal quote: ‘I stand on the shoulders of giants’ was attributed to Isaac Newton, as it turns out he wasn’t the one who coined it.
I can verify that I’m 5 and I do understand this thanks to this video.
I'm a mid/senior level webdev and I just come back to Alvin's videos when preparing for interview everytime. its so simple and help me rebrush all of them without much hassle. very good work
you probably use parent_id = .. db.Column in your projects, don't you xDDD
@@ordinaryggew.
@@ordinarygg Don't get it.
@@ordinarygg i know what u mean Xd, web devs isnt that hard 🗿🗿
I chose this video after I got an invitation for an online test from a well-known company. This was my only chance since I have never studied algorithm oriented programming.
I was given a graph problem and handled it correctly with a minor deficiency in efficiency :) Thanks a lot!
Was that FooBar? Did you finish it? They shut it down 😢
Was looking for something like this and can't beleive you just uploaded it!
Me too
This is the only course I’ve found that has successfully got the concept of graphs through my thick skull, it’s been a subject that’s been so hard for me to learn. Thank you for making this!!
“Later on in the tutorial we’ll go over examples of when u might use one over the other”
What a breath of fresh air 🙏🏼
the voice, he style of teaching , the animations and visuals absolutely phenomenal. love it
Man! you have my respect. This was the best course for getting started with graph problems I have ever seen. Thank You so much.
This is phenomenal! I was finding Graph problems so difficult before watching this video, and now they seem fairly easy after watching the entire video! Thanks a ton!
The course on DP was a winner. Course on Graph algo is another winner. Hope to learn a lot more from you.
you are another winner for appreciating him :)
@@keerthi1070 you are another winner for appreciating another winner
@@ziggystardust3763 you are a winner for appreciating a winner appreciating a winner
@@Tetrax this recursive call stack gotta stop somewhere XD
@@ldar6472 let’s assume we’re on an alternate universe where leetcode doesn’t have a time limit
How this person managed to explainamy concepts perfectly is superb... We really need people like this guy in the teaching field... I downloaded the video and I ve not regretted at all
Me too I just realized Breathe First Traverse is Fibonacci
This course is hands-down, unequivocally fantastic!!! Best useful course I've seen about graphs. It ties together the algorithms with concrete use cases which just clicked. Thank you for the fantastic job!!!!
This guy literally changed the way I look at dynamic programming.
This was amazing. I've watched a bunch of these types of videos over the years and this is the only one that actually made it look easy. I can actually say I understand this now. Thank you.
I never knew Graphs were that easy!
You're one of the rarest gems I have come across in my life.
Followed it all along. Solved all the problems in Python.
I'd always been scared of learning graphs, even though I wanted to... but this course took my fear away 😎👍🏻 excelente course, as always!
Just finished the Dynamic Programming from Alvin and now back at the Graph algorithm.
Leetcode - O(n^2) brute force.
Algoexpert - O(n) linear
Alvin / Structy - log(n)
This is a great intro to basic graph problems - however we may need a part two to cover more advanced topics such as Union Find, Minimum Spanning Trees, Dijkstras & bellman ford, topological sort, etc.
Sure we totally need. Code interview for wannabe juniors requires some of advanced topics
Guaranteed you won't actually need any of that when actually building something in the real world.
For island count (1:58:20), you actually don't need a visited set to keep track of the tiles you have already visited. Simply set the land tile to water "W" when you visit it; you already have the logic there to skip over water tiles.
Yes, but in real life, functions don’t modify input objects, it’s a bad practice. In an interview I wouldn’t do it without asking.
If space complexity needs to be minimized and the interviewer allows in-place modification, then this is a good approach.
OMG this video is an absolute gem! I used to be baffled by the island problem and just simply memorizing the solution on leetcode, but after following along for two hours, I can confidently code up the solution all by my self! Thank you so much you are a fantastic teacher.
Before watching this course, I dreaded DFS BFS problems. I always used to somehow understand the logic just before the interview and then forget everything. But now, I understand the logic, visualization and how to appraoch it. Thank you so much Alvin and FCC for this. I am definitely going to watch other interview videos. 👍
Thank you Alvin! I've been trying to grind leetcode and DSA for interviews, and this video as well as your binary trees video was so amazingly helpful and way more efficient than blindly doing problems to help me strengthen the fundamentals, and in particular, I was really struggling with handling how to handle graphs in the form of a matrix rather than an adjacency list -- I typically tried to convert it into an AL unnecessarily, but watching this video, especially the numIslands section, made things just click. You also have given me so much of a better understanding and less fear of recursion, even going to appreciation of recursion! That's crazy.
Also, for the numIslands problem, to save a bit on space, I got rid of the visited set and just set grid[row][col] == 'V' to mark that it's visited, and only add where grid[row][col] == L. This worked.
Huge fan of these courses, he's a great teacher and breaks down these problems to be far less intimidating
watched your lecture on Dynamic Programming and now I am here. You are such a phenomenal teacher, thanks a ton for making these tutorials!
Please @rahul what's the DP link
My guy, I don't know how to thank you. You have a marvelous way of teaching. This has really helped me a lot
Honestly one of the best tutorials ive come across in my 2 years of coding. Good work Alvin
Wow!, Your DP course was phenomenal, now you came up with Graph theory too, Thank you
Glad to hear you found value in my content! More coming soon.
@@AlvintheProgrammer what next ?
@@AlvintheProgrammer this is really true! you are the best!
@@vinayaktyagi8773 Likely trees or linked lists next. You can check out my platform, Structy (link in the video description), if you want to check it out early before it premiers on free code camp.
This is an incredibly straight forward and simple to understand graph video. This video definitely made me clean up rough edges in my understanding of tackling graph DSA problems. Thank you !!
50 mins in and THIS IS THE BEST VIDEO ON THE INTERNET handsdown!!
Thank you so much
totally agree!!!!
Best teacher ever. Thanks for this.
This is amazing. I found it really easy to understand the graph algorithms after referring to this video. Great job, Alvin :)
Small correction in the solution described for checking if a source and destination have a path in an undirected graph. (line no. 8)
The logic of traversing the neighbors and adding the source node to the visited list has to be part of the condition that validates whether ta node is visited or not.
if !(visited.has(src)) {
// add the src to the visitors list
// recursively traverse through the src neighbors
}
This course is exceptionally well done. Completely understood the theory and implementation behind them damned graphs.
Thanks
Thanks again, from Brazil! ❤
This was probably the most useful tutorial I've seen on YT. Great work!
Just for correction, Islands count problem, inside DFS function for checking boundaries, for 'j' check for j < grid[0].length, because if in NxM grid N !== M, in that case j < grid.length cannot check for true boundary and result will be wrong, I just wanna mention that in case
I was confused when you started going through depth first search because you put both of A's neighbors B and C on the stack which seems to be breadth first. But what helped me to understand is the fact that the loop starts when we pop a node from the stack, not when we add it to the stack. The stack doesn't represent which node we are currently "at", the node we pop from the stack does that. The stack is just an in between step. So we don't actually "go to" C at first, we just add it to the stack for now. We do "go to" B because we pop it from the stack
Your explanation is ideal. Voice and the way you expalin does not make a listener tired. This is just amazing. Please continue doing your job. It is amazing
I am actually studying C but I have found this video tremendeously helpfull in explaining how graphs are traversed and how not to use return in a bad way when you want to actually make use of recursion to backpedal from dead ends in situations where you don't find your base case.
thank you again. subscribed!
Best explanation I ever watched in UA-cam. You deserve a thumbs up and a comment from a laziest person on earth!
I never comment on UA-cam, but I love the way Alvin explains concepts. Thanks a lot for your work!
34:04 E = N*(N-1) for directed graphs. So in that example, 3 nodes means 6 edges (not 9, as in 3 squared)
This was an elite video. I did the last two questions and did not understand how to approach it or understood what the right answer for those questions. Watch the 2+ hours of video as well as coding it up myself, and everything makes sense. EVERYTHING. I am now able to do one of the toughest topics that I was having trouble with. So BIG Thank You for that.
P.S. For the last 2 questions, you don't need the visited set, you can flip the land to water or another value altogether. Avoiding the cyclic infinite loop.
Thank you for making graphs so easy for me! Appreciate your patience to make these videos and share
Never looked graph so easy to me before I watched this one, thanks! 😊
Thanks Alwin for this great tutorial. I had been trying to understand and get an answer to a similar set of problems for a very long time. You made my day, ...probably my life!
I am currently struggling with studying graphs, and you are my savior.
This is really really a complete tutorial for “Graph based Problems” both in terms of coding Problems, interview preperation including MAANG and even for a new bie.
So logical and calm and fantastic explanation. This is just unbeliveble :) How you can explain this stuff so easily. You are the legend
Changing lives, one algorithm at a time. Great work Alvin!
Here are the leetcode equivalent questions:
has path/ undirected path - 1971. Find if Path Exists in Graph
connected components - 323. Number of Connected Components in an Undirected Graph
island count - 200. Number of Islands
minIsland - 695. Max Area of Island
I like his lecture, learn a lots, thanks man.
I think this should be the first programming video anyone watches , I was struggling so hard before this
This course was incredibly helpful and so well done. Thanks so much! Alvin, you're a phenomenal teacher!!
As an embedded software engineer, I never learned about these kind of algorithms. This is invaluable knowledge for me, thanks for leveling up my skill!
easily the best teacher of algorithms I've ever seen. Alvin spends time going over theory, Big O notation, and common errors. As someone that's never studied algorithms and data structures, all the videos he has done have helped drastically
34:29 for anyone confused here, the graph in the example given is incorrect. number of edges is not n^2. To show a graph with n^2 edges each node needs to point to every other node AND also point to itself
Extremely high quality presentation! Well done
The best beginner graph tutorial.
I always come back to this to refresh myself on graphs.
Thank you so much. This is the only video on UA-cam which addresses the graph algorithms the way they should be addressed - graphical visualization, pseudocode and actual problem solving.
great video, but i have a minor comment
at 1:00:34, look at the hasPath function, i would swab the first 2 lines (line 7 and 8), which represent our 2 atomic cases, this way you can use the function to check whether there is a path from src to dst that does not pass through node x, or nodes {x, y, z}
by calling the function with the visited set already populated with the nodes you want to exclude, this would not work with the current implementation but it still works fine
Thank you so much, Alvin, I have always been intimidated by graphs. This is the best course you will ever stumble across on graphs.
This is fantastic. Makes graphs easy to digest. Thanks for that.
I did find a minor issue in the island count problem solution.
It assumes that each row will have the same number of columns. That assumption isn't described in the problem. So a quick fix would be to run the nested loop to the length of the row instead of fixing it to zero. And doing the same for the inbounds check. That would present an issue in the exploration in that there might not be any map in either direction, so you'll have to add an existence check in the explore to account for that.
This way if a row has less coordinates, or more, it would still work properly.
Preprocessing the input by storing edges in hashmap: this line took all my fears of not even trying to attempt graph problems to making me feel I could do any graph problems. Thanks for this video ❤️
Just have to agree with others that this is one of if not the best course on graph problems out there. Like many, I was intimidated by graph problems early on but this video has made me so much more comfortable with them. I feel like I actually understand them at a deep level now.
Awesome work! You inspired me to start my coding channel!
Great tutorial Alvin. For python programmers, here is the pattern for largest components (connected components is similar):
def explore(arr, node, visited, comp_count):
if node in visited:
return 0, visited
visited[node] = 1
for neighbour in arr[node]:
comp_count += 1
explore(arr, neighbour, visited, comp_count)
return comp_count, visited
def largestcomponent(arr):
largest_comp = 0
visited = {}
for node in arr:
if node not in visited:
comp_count, visited_new = explore(arr, node, visited, 1)
visited = visited_new
if comp_count > largest_comp:
largest_comp = comp_count
return largest_comp
Thanks to the first 30mins of this video alone I was able to extend the idea of adjacency lists and independently solve all the following problems on undirected graphs and grid graphs in just under a week's worth of practice
just wow on the fact that we're living in a world where we have access to such a quality learning material for FREE. Thanks a lot!!
by far the best explanation of dfs and bfs that actually makes sense!
This was not for coding interviews but a good start for beginners who are into graphs with perfect visualization
One of the most useful videos I've seen for understanding data structures, awesome
Quick observation on the Nondirected - Has Path problem:
Before that, Alvin, you are a genius man! You make everything sooo easy B)
Ok here it goes:
At around 59:00 you are using a set to keep track of visited nodes and returning false as soon as you find the source element in the set. I feel you are missing an edge case here.
Imagine a graph in the shape of a Kite 🪁🪁🪁🪁 --> A cycle on top and a tail just below the cycle. The cycle and the tail are one single graph.
Let us say the cycle starts at node a (a - b - c - a) and the tail (a - m - n - o). I want to find a path between (b and o)
We need to to track the remaining nodes of the graph, because, without it, the process will start from a and take a trip through the cycle and come back to a; this already exists in set at this point in execution, the function returns false without checking the tail section of the Kite Shaped Graph.
Here because Alvin's DP video was amazing. This is another gem. Thanks!!!
Learning this to win a coding competition. If I reply to this thread on the 28th of this month, it's good news! Wish me luck🙏
I am guessing that's not a good news! Hope you are doing well.
¡Gracias!
this guy is the best teacher there is.
Thank you so much!
I have subscribed to your chanel (:
100% the best teacher I've seen. You are legit the best, my dude.
I thought I'd seen the all the greatest of online coding instructors, and then I found Alvin... Seriously these explanations are paced exactly perfectly and ultra clear
amazing tutor, The best I've seen
One issue in the shortest path, 1:36:16 , its that whenever you find the destination you always return it. Its not always the shortest path.
You can maintain a min counter to maintain the shortest distance.
Also thanks for the wonderful video, just by following the videos , I could do the implementation on my own
I think that would have been an issue if we were using DFS, considering the nature of BFS and visited set that we are using, whenever we find the destination, that path will always be the shortest.
@@siddhantgupta4295 correct, thanks for the detailed explaination. I came to know about this after doing many bfs problems. I dont knw if it was mentioned in this video, if not then it should be mentioned :) happy coding
Love you man. You make things so easy. Please do more courses. Will support you to the moon.
You just made me crack my Interview on graph algorithms... and I have just seen half the video only to make it... Thanks a ton Bro... Let me complete this
number islands problem couldnt have been solved easier than this, kudos to this guy! for once i was able to understand the solution and complete the problem without errors.
You're doing an amazing job Alvin👍I'm always on lookout for your contents. Thank you!
For the island count problem, i ended up converting the entire 2D grid into a graph containing only 'L' nodes with multiple components. I then did a dfs for each of those islands, incrementing a counter everytime i finished exploring one of them and adding it to my visited set. Really interesting how you can solve this problem in such a different way!
Nice try. You converted the problem to components based & followed his components count approach.
Finished my very first fCC course. The pedagogy is excellent, Alvin is a great teacher
I was quite struggling with problems when I realized my conception isn't clear. I can't thank you enough for this explanation.
I literally have no words to express how mind-blowing Alvin is.
Hats off FCC and Alvin. Whenever I see your videos I also get the feeling that I should give back to the awesome community this is .
For largest component problem , there can be an easy and alternate solution . Since , we already have visited which is keeping track of all nodes. There is no need to maintain separate "size". Below is the code you can use to your advantage:
const largestComponent = (graph) => {
let maxSize = 0;
let previousVisitedSize = 0;
let actualSize = 0;
const visited = new Set();
for(const node in graph) {
if(explore(graph, node, visited) === true) {
actualSize = visited.size - previousVisitedSize;
maxSize = Math.max(actualSize, maxSize);
previousVisitedSize = visited.size;
}
}
return maxSize;
};
const explore = (graph, node, visited) => {
if(visited.has(String(node))) return false;
visited.add(String(node));
for(const neighbor of graph[node]) {
explore(graph, neighbor, visited)
}
return true;
}
The series of your videos are amazing. They are the best courses I have ever seen regarding data structure and algorithm studies. However, I do think I find out a potential mistake. At 1:53:55, check the line 12, const colInbounds = 0
All of Alvin's content is amazing! I signed up for Structy after doing this course and the course on trees, and love it!
Thank you for the video. Also to find the minimum sizes of the island, we can use the method that calls the recursive function (parent method) to hold an integer that points to the size of the visited set before the recursive call and then use the difference in size after the recursive call to determine the minimum size. This to me didn't require me to create a new recursive method that returns an integer. However, creating an extra method does have its benefits too.
Thank you Alvin! Your explanation is GOLD!!! It's such a phenomenal tutorial that I find myself learning so much more efficiently!
This is really a damn refresher
Alvin is single-handedly going to take me to the promised land. Love you Alvz
This video is game changer!!