Whenever I struggle with daily Leetcode problems, I turn to Neetcode

@shreehari2589

3 ай бұрын

Ok

I was only able to implement a brute force solution, but only 67 out of 71 test cases passed (got time limit exception). 😁Thanks for the video.

Thanks for this. Please do daily LCs and Contest please.

Thanks for putting this up!

Thank you, I needed this

really cool stuff man, thanks for the content

Thank you so much for the daily. Really helps a lot.

Thanks for this

Perhaps slightly simpler (more familiar BFS) impl where we can get rid of the layer traversal technique. At the end of the while loop, all root will have the highest layer value (since they're the last layer) 😃 // Initialize layers with -1, except leaves layers are 0. int max_layer = 0; while (!q.empty()) { int u = q.front(); q.pop(); link[u]--; for (int v : adj[u]) { link[v]--; // Here, one of u's neighbors v just turned into a leaf node after u is removed. if (link[v] == 1 && layers[v] == -1) { layers[v] = layers[u] + 1; max_layer = max(max_layer, layers[v]); q.push(v); } } } vector res; for (int i = 0; i if (layers[i] == max_layer) { res.push_back(i); } }

I solved it using re-rooting. But this "Removing leaf nodes" solution is very amazing. I didn't thought of that. Thank you so much!💖

Thanks 😊

Missed you buddy 😭

Hey Navdeep how are you? I love your leetcode explanations and your solutions! Gives me motivation to do more leetcode

Good one

edging to this rn

Kahn's algorithm as intuition for this question. topsort(remove) where indegree ==1

I liked the note about the length and how other languages would calculate the length. With so much python sugar, we forget how a similar code would break in Java or C.

Great video as always! One thing that's worth mentioning is that if n

@KaranBulani

3 ай бұрын

was thinking same

@bellxlilies9913

3 ай бұрын

Yes, if the n

code: class Solution: def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]: if n==1: return [0] adj = defaultdict(list) for n1,n2 in edges: adj[n1].append(n2) adj[n2].append(n1) edgecount = {} leaves = deque() for s, nei in adj.items(): if len(nei)==1: leaves.append(s) edgecount[s]=len(nei) while leaves: if n

it takes me over 15 minutes to be clear the requirement,and until i finished the solution, it takes more than 1 hour.

@pixusru

3 ай бұрын

You’re not supposed to come up with requirements and approach on the spot. You must know every possible problem or pattern by heart, because many do, so you’re looking bad compared to them. Sorry, that’s the game.

@kumarc4853

3 ай бұрын

@@pixusru true, the people who post this arent solving this live and for the first time either. So solve as many as you can and spot patterns, recognize general strategies, things to think about when blocked etc

@sophiophile

2 ай бұрын

​@@pixusru Tell that to interviewers at FAANG companies :(. I have spent a long time preparing for the code portion of an interview w/ one tomorrow (even though I already have another role essentially secured as a backup). There's always going to be the chance that you just don't spot the optimal solution at that time, no matter how prepped you are. For example, most FAANG would not accept the solution coded up here and would require the second approach he described (since it is O(n) time, O(1) memory- while the first is O(n)/O(n))

For the edge case, you could check if length of neighbours is 0 or 1, then it will pass the edge case, without having to create an seperate check for the edge case

No need to check if n==1: return[0], we can return [0] after the while loop and it will work fine.

Crazy hard

Okay. At first I thought it was just BFS from the leaves until you have 2 or 1 left. Lots of tests failed. The whole subtracting edges thing went over my head. Second explanation was easy to get though I haven't tried it yet.

This question was super hard, first I tried solving it with DP with memoization, it gave me TLE on test 70 I think

Can you do a video using Morris traversal?

The brute force solution is still hard to do

Is it ok if after watching an entire video, didn't get the solution considering I only have a basic knowledge of graphs and DFS

This problem should be renamed from "Minimum Height Trees" to "Find the root". We can imagine like cutting leaf from all sides equally at same level. What is left is root ( can be two node or one ).

The second approach feels more intuitive for me...

Is the time complexity of the code O(n) ?

@amitchoraria5737

3 ай бұрын

i think yeah should be O(n). but just want someone to confirm

@juanito1410

3 ай бұрын

The algorithm here is called khan's algorithm and its TC will be O(V+E) where V is the number of vertices/nodes and E for the count of edges.

omg i hate graph problems, I hope one day they become less intimidating

Proof why leaf nodes can't be a root of min path. Suppose exist a min path, where leaf node is a root A(leaf) - B(end) , the node A has one neighbor N and this neighbor has to be included into the path A-B , from this we can find a root with smaller path N-B

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