🚀 neetcode.io/ - A better way to prepare for Coding Interviews
@jchakrab
2 жыл бұрын
think of a directed graph 0->1->2->3->4, isn't your solution's time complexity O(E * N**2)...you will start the same loop for 1, 2, 3, 4 after doing it for 0
@xmnemonic
Жыл бұрын
The .remove(crs) was so confusing but I finally understood it. Simplest explanation: if we exit the for-loop inside dfs, we know that crs is a good node without cycles. However, if it remained in the visited set, we could trip the first if-clause in the dfs function if we revisit it and return False. That's what we don't want to do, because we just calculated that crs has no cycles. So we remove it from visited so that other paths can successfully revisit it. Basically we can visit the node twice without it being a cycle due to it terminating multiple paths.
@yashjakhotiya5808
Жыл бұрын
and the reason it terminates at 'twice' is because of preMap[crs] = []
@tiffanychan6272
Жыл бұрын
thank you! i was pulling out my hair trying to figure out why
@wayne4591
Жыл бұрын
Actually you can see this trick in many graph, binary tree or other problems using back tracking. Because the visit set is only used to contain the current visit path. So whenever you exit the sub-function you create on this level, you have to pop the info you passed in before.
@sidazhong2019
Жыл бұрын
in every dfs, pop() or remove() after a call, is a standard process. you will see.
@netraamrale3850
Жыл бұрын
Do you have any example?
@juliahuanlingtong6757
3 жыл бұрын
The setting of preMap[crs]=[] before return true is so smart!!! Absolutely love it
@yuemingpang3161
2 жыл бұрын
Pretty smart! He removes all pre-courses at once after iterate through one adjacency list. In the drawing solution, he removes the pre-courses one by one. To align with the codes, the list can only be "cleaned out" if all pre-courses returns true. Actually, I was a little bit confused when I first saw the codes.
@alfahimbin7161
Жыл бұрын
@@yuemingpang3161 what is the time and space complexity of this solution??
@yashjakhotiya5808
Жыл бұрын
And necessary for time complexity to be O(num_nodes). If we didn't do it, the last for loop would have us visiting nodes as many times as it required by other courses, making the overall complexity O(n^2).
@minepotato7126
5 ай бұрын
It's too smart
@EranM
3 ай бұрын
not only smart, but essential for a good running time.. hell I fell there!!!
@saralee548
2 жыл бұрын
Your channel is soo helpful! If I get stuck on a LC question I always search for your channel! Helped me pass OAs for several companies. Thank you so much.
@idgafa
Жыл бұрын
If you move the line 13 `if preMap[crs] == []` before the line 11 `if` check, then you don't need the `visitedSet.remove(crs)` in line 19, because you will never traverse the visited path that way. Thanks for great explanation.
@Alex-tm5hr
Ай бұрын
You can actually just remove it all together and it still passes lol
@yoshi4980
3 жыл бұрын
this is a clever solution. not something i would ever come up with haha, i had a similar idea, but it kind of just broke down during the dfs step. i had a lot of trouble trying to figure out how to detect cycles in a directed graph...in the end when i was looking in the discussion i saw that you could just do a topological sort so i felt silly after that haha. gotta work on graph problems more :-)
@alfahimbin7161
Жыл бұрын
what is the time and space complexity of this solution??
@frida8519
Жыл бұрын
@@alfahimbin7161 O(n + p) where n is the number of courses and p the prerequisites. The explain it at around 08:37
@saumyaverma9581
3 жыл бұрын
He is speaking the language of god.🔥🔥
@DarkOceanShark
2 жыл бұрын
Thank you so much pal! I was able to crack it myself after seeing your visualizations of the graph, with ease. Words can't describe my happiness.
@alfahimbin7161
Жыл бұрын
what is the time and space complexity of this solution??
@momentumbees3433
2 жыл бұрын
To simplify this problem This is based on finding if the directed graph has a cycle. If yes then return false(cannot complete all courses) else return true.
@tonyiommisg
5 ай бұрын
I feel the way the problem is worded, I never realized that it was asking this question lol
@devdoesstuff
8 күн бұрын
Here is a simpler solution. Instead of removing and adding from a map, we simply use a status 0 -> unvisited, 1-> visiting, and 2->visited for tracking the current node status in a DFS path. If the node is revisited before completing all neighbors, it would have status 1 which implies there is a cycle in the graph. Same logic as Neetcode's though. class Solution: def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: graph = [[] for _ in range(numCourses)] visit = [0]*numCourses for u,v in prerequisites: graph[u].append(v) def dfs(node): if visit[node] == 1: return False if visit[node] == 2: return True visit[node] = 1 for neighbor in graph[node]: if not dfs(neighbor): return False visit[node] = 2 return True for i in range(numCourses): if visit[i] == 0: if not dfs(i): return False return True
@steffikeranranij2314
3 жыл бұрын
What a lucid explanation! Keep this up!
@alfahimbin7161
Жыл бұрын
what is the time and space complexity of this solution??
@sanaa3151
2 жыл бұрын
this was so so helpful, thank you so much for being so clear!
@NeetCode
2 жыл бұрын
Glad it's helpful!
@tuandino6990
10 күн бұрын
I find using topological sort for this task much more intuative and easier to implement
@rahul911017
2 жыл бұрын
I have taken Neetcode's Course Schedule 2 idea and implemented this on those lines: Personally I found that idea more intuitive and easier to follow. class Solution { HashMap prereq = new HashMap(); // hashset to mark the visited elements in a path HashSet completed = new HashSet(); // we use a hashset for current path as it enables quicker lookup // we could use the output to see if the node is already a part of output, // but the lookup on a list is O(n) HashSet currPath = new HashSet(); public boolean canFinish(int numCourses, int[][] prerequisites) { // base case if (numCourses
@xingdi986
3 жыл бұрын
If you want to take course 1, you have to take course 0 first.
@yynnooot
3 ай бұрын
I was really confused about the direction of the edges. Intuitively, I would think precourse -> course, but you have the arrows going backwards from course -> precourse. By switching the arrows around to: precourse -> course, and having your adjacency list as: { precourse: [ course ] } instead of: { course: [ precourse ] }, your DFS solution still works. The benefit to doing it this way is that you can use the same adjacency list pattern for a BFS topological sort approach, which needs access to the neighbors of nodes with zero in-degrees.
@JoffreyB
2 жыл бұрын
You draw edge incorrectly. If it's [0, 1] meaning you first have to take course 1 before 0, edge is gonna be 1->0, not 0->1, because first we need to take 1 and only then we will have access to the 0.
@chrisgeorge2420
2 жыл бұрын
his solution models the graph in the other direction, it is still correct because he is consistent with it
@yynnooot
2 жыл бұрын
I was thinking the same thing, the arrows threw me off
@mostinho7
2 жыл бұрын
The arrows/wording is messed up, he’s using the word prerequisite to actually mean postrequisite, but good video still
@lemonke8132
2 жыл бұрын
yeah all his arrows are backwards I don't know how that makes sense to him.
@Captainfeso
3 жыл бұрын
Thanks for the very clear explanation. I have a suggestion for direction of arrows that may be less confusing. For example, prerequisites = [[1,0]] means that if we have to take course 0 before course 1. So my graph would be pictured like: 0------->1 instead of 1------>0.
@ua9091
3 жыл бұрын
Depends on how we see it. In his case, the concept is like 1 has a dependency on 0, hence drew an edge from 1 pointing towards 0.
@halahmilksheikh
2 жыл бұрын
Yeah that was so confusing
@ax5344
3 жыл бұрын
I have a hard time envisioning visited.remove(crs). I cannot connect this to the "Drawing Solution" earlier. I can see when preMap[crs] is set to 0 @7:44, but I cannot see any part where visisted.remove(crs) corresponds to. I understand to detect a cycle, we need to visisted.add(crs), But I cannot see where visited.remove(crs) fits. Can someone help?
@RanjuRao
3 жыл бұрын
Lets say course 3 is dependent on 2 and 2 is on 1, while traversing for 3 you make dfs(2) which in turn is dfs(1) but dfs(1) does not have pre-req so u mark it as [] (initially) and similarly you need to mark dfs(2) to [] which is done using set.remove() and map.append([] ) for key =2 .
@akinfemi
3 жыл бұрын
Same. What helped me was thinking about it as setting the course node to a "leaf node". If you notice the leaf nodes (courses with no prerequisites) are never added to the visited set. So setting it to [] and removing it from the set does that.
@sudluee
3 жыл бұрын
I think it makes more sense if you replace visitSet.remove() with visitSet.clear(). visitSet.clear() also works for our purpose, which is basically to give us a new visitSet for each course so we don't get false positives from earlier runs.
@jessepinkman566
2 жыл бұрын
When the graph is not fully connected. 1->2->3, 4->3. If you should not remove 3, 4->3 would be false because 3 is already in the set. However, you can also choose to change the order of the two ifs in the start of the bfs to avoid removing.
@tonyz2203
2 жыл бұрын
feel the same thing.
@farleylai1102
2 жыл бұрын
LintCode imposes the memory constraint such that recursive DFS will fail. The intended solution should be iterative based topological sort.
@MinhNguyen-lz1pg
2 жыл бұрын
Great explanation, I was doing the adj list pre->course and confused myself in the coding step. Thanks for the video, smart ideas. I definitely was not thinking of the fully connected graph case
@pat777b
Ай бұрын
I implemented DFS via a stack. I also tried a similar approach with BFS using a deque queue but it was a lot slower. class Solution: def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: graph = defaultdict(list) for prereq in prerequisites: graph[prereq[0]].append(prereq[1]) for i in range(numCourses): if i in graph: stack = graph[i] seen = set() while stack: node = stack.pop() if node == i: return False if node not in seen: for j in graph[node]: stack.append(j) seen.add(node) return True
@vivekshaw2095
2 жыл бұрын
you dont need to remove crs from visited at the end if you just check adj==[crs] before checking crs in visitSet
@user-j5ja95
11 ай бұрын
huh ?
@giraffey8
3 жыл бұрын
I literally looked at this question yesterday and couldn’t get it, thanks for making this vid!
@NeetCode
3 жыл бұрын
A nice coincidence! Thanks for watching
@ashleal5655
3 жыл бұрын
Java version of this solution: class Solution { Map preMap = new HashMap(); Set visitSet = new HashSet(); public boolean canFinish(int numCourses, int[][] prerequisites) { for (int i = 0; i < numCourses; i++) { preMap.put(i, new ArrayList()); } for (int[] p : prerequisites) { List neighbors = preMap.get(p[0]); neighbors.add(p[1]); preMap.put(p[0], neighbors); } for (int i = 0; i < numCourses; i++) { if (!dfs(i, preMap, visitSet)) return false; } return true; } public boolean dfs(int course, Map preMap, Set visitSet) { if (visitSet.contains(course)) return false; if (preMap.get(course).size() == 0) return true; visitSet.add(course); for (int pre : preMap.get(course)) { if (!dfs(pre, preMap, visitSet)) return false; } visitSet.remove(course); preMap.put(course, new ArrayList()); return true; } }
@tamilcodingclub3832
2 жыл бұрын
you saved me! Thankssss!
@vsbgugan
2 жыл бұрын
@@tamilcodingclub3832 Instead of preMap.put(course, new ArrayList()); you can do preMap.get(course).clear(); It saves memory
@eliasl7705
2 жыл бұрын
Thank you so much for all of these videos. Very well explained and also well put together and displayed. Really fantastic material, it's been absolutely invaluable in helping me to learn and improve my skills.
@chloe3337
2 жыл бұрын
Could you also go through the space complexity in your videos?
@pacomarmolejo3492
Жыл бұрын
Given N = number of courses, P = prerequisites; TC: O(N + P), because we are visiting each "node" once, and each "edge" once as well. SC: O(N+P), as our hashmap is of size N + P, and the recursive call stack + visited set are of size N.
@Allen-tu9eu
3 жыл бұрын
you are the very best one for explain leetcode problems, and I am not even a python user
@NeetCode
3 жыл бұрын
Thanks! =)
@no3lcodes
Жыл бұрын
Intuitively I was thinking of this solution and gave myself 30 mins to solve it, I got to the part of DFS but then confused myself because I was like "I feel like I need DFS here but where should I start it, how should I call it" and then the time was up xD, I'm happy I almost came up with it alone though. Thanks for the video, it clarified what I was having trouble with.
@jritzeku
5 ай бұрын
Explanation on WHY/HOW cycle was detected(The crux of the problem) -As we perform dfs, we add node to 'visited' set if it does not exist. -Once we have exhausted all its neighbors/prerequisites AND return back to it from call stack, we pop it from call stack and we remove from 'visited' set. -A cycle is detected when the node we are popping off of call stack still exists in 'visited' set. BUT WHY?? In the last example he provides @ 10:50, while we have/are visiting the last neighbor/prepreq of node 0, unfortunately we have not returned back to it due to its order in call stack. Hence a cycle was detected before we we're able to remove node 0 from call stack.
@TheLaidia
3 жыл бұрын
clear solution, thank you! Wish you could also go over BFS 😄
@yumindev
2 жыл бұрын
in the example case [1,0], why is it out reach arrow 1-> 0 ? I thought in order to get 1, you need to get 0 first, so, it's 0->1 ? am i right? it's a little anti-intuitive ?
@msm1723
2 жыл бұрын
@NeetCode Thank you so much for your work! I'am going through collection of your solutions and litterally feeling smarter) I don't really understand one thing in this problem - why do they provide numCourses at all? I mean, when given number is less then total number of courses provided in prerequisites - like (1, [[0, 1]]) the algorithm fails. And of course you dont realy need this number to create preMap (you could use defaultdict, or check if key exists on string 14 before comparing to empty list). Iteration through length of preMap also will work when running dfs.
@wayne4591
Жыл бұрын
Thanks for your explanation! it is clean and easy to percept. I really appreciate your coding style. Just a heads up that the if perMap[crs] == [] return True can be omitted since if we have an empty array for prerequisites for crs, the for loop afterwards will just end and return True at the end!
@pacomarmolejo3492
Жыл бұрын
Yes, though, you save "some time" by handling it that way.
@vijethkashyap151
3 ай бұрын
I feel using Kahn's algorithm for detecting cycles in the directed graph is the simplest solution for this, even though logically not 100% right as we use indegree in Kahn's algorithm, as soon as I see cycle and undirected graph I solved it with this method. Then I got to know we are supposed to deal with outdegree to deal with independent nodes in this case! Though Kahn's algo works !
@beaglesnlove580
2 жыл бұрын
Hey man, thanks a lot for your description. You probably have the best explanation on this this problem compared to other KZitemrs. There’s a girl that’s pretty good too her names jenny
@varshard0
Жыл бұрын
When I did this exercise for the first time, I actually created a whole complete graph data structure from scratch. Then created 2 visited maps to resolve the circular issue. So much memory required
@beksultanomirzak9803
2 жыл бұрын
You explanaition is amazing, I love it !
@NeetCode
2 жыл бұрын
Glad it's helpful!
@meylyssa3666
3 жыл бұрын
Thank you! Great explanation! And you have a magical voice, such a pleasure to listen to your explanations.
@RandomShowerThoughts
4 ай бұрын
without the empty list optimization this will get TLE, pretty smart to figure that out
@melvin6228
3 ай бұрын
I tackled this problem as cycle detection.
@qx5234
2 жыл бұрын
Your explanation is super clear, thanks
@alfahimbin7161
Жыл бұрын
what is the time and space complexity of this solution??
@user-ik4ju3vs2z
10 ай бұрын
This is very clear explanation, but I met Time Limit Exceeded problem, so I made the follow changes to met the requirements: class Solution(object): def canFinish(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: bool """ adjacency_list = [[] for _ in range(numCourses)] for crs, prec in prerequisites: adjacency_list[crs].append(prec) visited = [0] * numCourses def dfs(crs): if visited[crs] == 1: return False if visited[crs] == 2: return True visited[crs] = 1 for prec in adjacency_list[crs]: if not dfs(prec): return False visited[crs] = 2 return True for crs in range(numCourses): if not dfs(crs): return False return True
@dcc5244
6 ай бұрын
The more you learn about recursion, the more crazy you become.
@chenyu-jg4kg
7 ай бұрын
Brilliant Solution!! It took me a while to think it through but finally understood it, really appreciate your help!
@tonyiommisg
5 ай бұрын
I have to say if you didn't realize this was a graph problem or to just think about it literally initially about courses and prerequisites you can get pretty stuck with where to go. :(
@jakubucinski
2 ай бұрын
Simpler version (imo): class Solution: def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: pre = collections.defaultdict(list) for e in prerequisites: pre[e[0]].append(e[1]) visited = set() completed = set() def dfs(node): if node in visited and node not in completed: return False if node in visited: return True visited.add(node) res = all(dfs(child) for child in pre[node]) completed.add(node) return res return all(dfs(n) for n in range(numCourses))
@holdeneagle7734
3 ай бұрын
Forgot the remove part. You are amazing
@yuchenzhang1741
3 жыл бұрын
SO clear! Thanks a lot
@zhoudavid450
3 жыл бұрын
I meet this question today, thank you so much.
@dorondavid4698
2 жыл бұрын
In case others come across this, these types of questions are classified under Topological Sort as well
@zr60
2 жыл бұрын
He's using a hacky method instead of topological sort.
@PippyPappyPatterson
2 жыл бұрын
Any resources on Topological Sort?
@PippyPappyPatterson
2 жыл бұрын
Or, do you have any leetcode problems and solutions that you've implemented using topo sort?
@surajpasuparthy
2 жыл бұрын
we dont need to travese 4 from 1 if we are keeping track of the visited nodes. cross edge iterations can be eliminated to increase the speed of the algorithm, right?
@DataStructures
Жыл бұрын
this question is currently being asked at Amazon. My brother is one of the interviewers who asks it hehe
@pigeonfanatic13
2 жыл бұрын
you can use a defaultdict(list) instead for preMap :)
@MorbusCQ
2 жыл бұрын
Hello everyone, this is YOUR daily dose of leetcode solutions
@Rahul-pr1zr
3 жыл бұрын
Nice explanation. Curious - why/how did you zero in on using DFS instead of BFS?
@hillarioushollywood4267
2 жыл бұрын
@rahul, to check if a particular course completion can be possible. And we can do it if and only if we can check all its prerequisite.
@MP-ny3ep
Жыл бұрын
Phenomenal explanation! Thank you so much!
@glife54
7 ай бұрын
thanks for the n = 5 expample, cleared the ques for me !
@obesechicken13
Ай бұрын
I think you have it backwards around 1:19 but no big deal. 1 is a prereq of 0
@mangofan01
8 ай бұрын
Bro, you are just GOLD!
@timmyzsearcy
2 жыл бұрын
In the beginning you show the edge going the wrong way for [1,0] the direction of the arrow should be from 0 to 1
@safakozkan6698
10 ай бұрын
Lines 13 and 14 are redundant. if preMap[crs] == [], code will skip the for loop and return True at L 21 already
@mikedelta658
6 ай бұрын
Killer explanation. Thank you.
@ayzchen1
Жыл бұрын
Thank you for this awesome video! I am wondering if you could do a video about a BFS version of the same problem? Thank you very much!
@ptreeful
Жыл бұрын
I don’t quite understand. Is it about topological sort or some other kind of algorythm? Like finding cycles for example
@rohit-ld6fc
2 жыл бұрын
so isnt it just a detect cycle problem? if cycle exists return false else true ?
@a2xd94
Жыл бұрын
Is it just me or does it seem like Leetcode has added a testcase for this problem that causes it to exceed time limit, even with this great implementation? Cannot get a pass for this problem...
@chakpang147
Жыл бұрын
Did you set the dependencies list to empty after a node is searched? This would save a lot of time
@jxw7196
Жыл бұрын
Very nice. Thanks for making this
@tyler5244
Жыл бұрын
Viewed this solution after an hour of trying to solve it and yep like usual there's no way my dumbass self could come up with this on my own
@ruthylevi9804
2 жыл бұрын
Love your videos! Would love to see the code included as well, especially in Javascript as converting can be tough. Thanks NeetCode :)
@fazliddinfayziev-qg1vg
10 ай бұрын
So amazing brother. Thank you
@N.I.C.K-
2 жыл бұрын
Thank you!!! Such a great teacher
@sagardafle
2 жыл бұрын
Thanks Neetcode! Is this playlist supposed to be followed sequentially? Thanks
@niteshmanem1997
2 жыл бұрын
neetcode aka beastcode strikes again with a flawless solution. LETS GOO!!!
@jackwilliamson2943
Жыл бұрын
The way you set up the edges is very unintuitive. I think of it as if 0 is a prerequisite for 1 then 0 -> 1. So you'd traverse the graph in the order you would take the classes. Otherwise good video!
@briansyoutubechannel8516
25 күн бұрын
Am I the only one who is confused why numCourses is even part of this problem? All the test cases in the leetcode are either cyclic (output: false) or non cyclic (output: true), so this problem can be solved with Kahn's algorithm.
@EranM
4 ай бұрын
preMap = defaultdict(list)
@c0dertang
Жыл бұрын
"because i like it better" - Neetcode 2021
@chloexie6576
2 жыл бұрын
good explanation on dfs, thanks! and i like the name of your channel :)
@NeetCode
2 жыл бұрын
Happy it's helpful :)
@confused_Creator_
2 жыл бұрын
Wow... The best explaination... ❤️
@dhaanaanjaay
4 ай бұрын
You are legend man!!!
@shuyangnie2446
2 жыл бұрын
Thank you neetcode guy
@anushayerram772
Жыл бұрын
@NeetCode Can you please explain the time complexity in detail.
@TheIcanthinkofaname
2 жыл бұрын
Awesome solution!
@bombdotcomist
Жыл бұрын
Brilliant explanation, wow.
@briangurka8085
3 жыл бұрын
Dude this literally SAVED me!
@kanchankrishna3686
2 жыл бұрын
Why can't your map be an integer as the key and an integer as the value? Why does the value have to be a list? I thought it would be okay to declare the map as int, int since you have one prereq for each course
@hikemalliday6007
Жыл бұрын
this channel is great
@denisgabrielcraciun
2 жыл бұрын
Could another solution be just detecting if the graph contains a cycle or not? You would use 2 pointers and make one move by one position and the other by 2. If the nodes encounter each other then you will have a cycle therefore you can’t complete the courses. Could someone tell me if there is a flaw in this solution? Thanks :)
@d4ntoine134
Жыл бұрын
You wouldn't detect isolated courses
@JuanGonzalez-cl2fy
3 жыл бұрын
Thanks for this wonderful video!
@NeetCode
3 жыл бұрын
Thanks!
@danielsun716
2 жыл бұрын
I still don't know why we should do "visitSet.remove(crs)". Neetcode said cause we do not use it. But why? can't we still leave it there? why not?
@gokulnaathbaskar9808
2 жыл бұрын
Thank you so much!
@sabinbajracharya3815
2 жыл бұрын
Solution: Without building a PreMap: - visitedLocal is used to detect the loop - visitedGlobal is used track if we've already been through the course. fun canFinish(numCourses: Int, prerequisites: Array): Boolean { val visitedLocal = Array(numCourses) { false } val visitedGlobal = Array(numCourses) { false } fun dfs(csr: Int): Boolean { if (visitedLocal[csr]) return false if (visitedGlobal[csr]) return true visitedLocal[csr] = true visitedGlobal[csr] = true for (p in prerequisites) { if (p[0] == csr) { if (!dfs(p[1])) { return false } } } visitedLocal[csr] = false return true } for (i in 0 until numCourses ) { if (!dfs(i)) { return false } } return true }
@codedoctor3265
3 ай бұрын
BFS , More intuitive solution class Solution(object): def canFinish(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: bool """ graph = defaultdict(list) in_degree = [0] * numCourses for course, prereq in prerequisites: graph[prereq].append(course) in_degree[course] += 1 # Step 2: Initialize the queue with courses having in-degree of 0 queue = deque([i for i in range(numCourses) if in_degree[i] == 0]) # Step 3: Process the courses using BFS processed_courses = 0 while queue: current_course = queue.popleft() processed_courses += 1 # Reduce the in-degree of neighboring courses for neighbor in graph[current_course]: in_degree[neighbor] -= 1 # If in-degree becomes 0, add it to the queue if in_degree[neighbor] == 0: queue.append(neighbor) # Step 4: Check if all courses are processed return processed_courses == numCourses
@szzz32112
2 жыл бұрын
much clear than leetcode sol
@namelesslamp12
3 жыл бұрын
looking sharp thanks
@sugalump9749
Жыл бұрын
this is actually a cycle detection problem. I have no idea why the problem was framed as an adjacency graph problem
@julianelmasry9556
Жыл бұрын
Is this considered topological sort?
@theodoretourneux5662
11 ай бұрын
how come you don't use a set instead of a list for the visitedSet? One would need to use the nonlocal keyword but the lookup times are much quicker. Couldn't there be an edge case where your last node in the dfs loops back to the second to last and then you are searching the whole array. This could potentially happen a couple times no? Thanks for any clarity you can provide!
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